Are there solutions to $a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ in positive integers $(a,b,c,d)$ with $d>1$?
0
$begingroup$
Let be $a, b, c, d$ positive integers.
Consider the following equation:
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
$(a,b,c,d)=(1,1,1,1)$ is a solution
$(a,b,c,d)=(5,4,6,1)$ is another solution.
Are there integer solutions with $d>1$?
number-theory
edited Jan 5 at 18:21
Blue
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
$endgroup$
add a comment |
0
$begingroup$
Let be $a, b, c, d$ positive integers.
Consider the following equation:
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
$(a,b,c,d)=(1,1,1,1)$ is a solution
$(a,b,c,d)=(5,4,6,1)$ is another solution.
Are there integer solutions with $d>1$?
number-theory
edited Jan 5 at 18:21
Blue
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
$endgroup$
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33
add a comment |
0
0
0
1
$begingroup$
Let be $a, b, c, d$ positive integers.
Consider the following equation:
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
$(a,b,c,d)=(1,1,1,1)$ is a solution
$(a,b,c,d)=(5,4,6,1)$ is another solution.
Are there integer solutions with $d>1$?
number-theory
edited Jan 5 at 18:21
Blue
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
$endgroup$
Let be $a, b, c, d$ positive integers.
Consider the following equation:
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
$(a,b,c,d)=(1,1,1,1)$ is a solution
$(a,b,c,d)=(5,4,6,1)$ is another solution.
Are there integer solutions with $d>1$?
number-theory
number-theory
edited Jan 5 at 18:21
Blue
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
edited Jan 5 at 18:21
Blue
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
edited Jan 5 at 18:21
Blue
49.1k870156
edited Jan 5 at 18:21
Blue
49.1k870156
edited Jan 5 at 18:21
Blue
49.1k870156
49.1k870156
asked Jan 5 at 18:03
user631773user631773
11
asked Jan 5 at 18:03
user631773user631773
11
asked Jan 5 at 18:03
user631773user631773
11
11
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33
add a comment |
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33
add a comment |
4 Answers
4
active
oldest
votes
1
$begingroup$
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
$endgroup$
add a comment |
1
$begingroup$
It seems that $frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :
(1,1,1,1)
(2,2,2,2)
(3,2,2,4)
(3,3,3,3)
(5,3,5,6)
(8,3,9,6)
(4,4,4,4)
(6,4,4,8)
(5,5,3,6)
(5,5,5,5)
(6,6,6,6)
(7,7,7,7)
(8,7,9,4)
(8,8,8,8)
(8,9,3,6)
(8,9,7,4)
(9,9,9,9)
answered Jan 5 at 18:22
YoshiYoshi
133
$endgroup$
add a comment |
0
$begingroup$
I get it now. Begin with any triple such that $gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy.
$$ k = frac{2b^2+2c^2+d^2}{gcd left( 2b^2+2c^2+d^2 ;, ;2b^3+2c^3+d^3 right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
0
$begingroup$
For the equation.
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
You can record this parameterization.
$$a=p^3-3s^3$$
$$b=p^3+sp^2-9s^3$$
$$c=p(p^2-6s^2)$$
$$d=p^3-2sp^2+9s^3$$
answered Jan 6 at 9:03
individindivid
3,2621916
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062994%2fare-there-solutions-to-a-frac2b32c3d32b22c2d2-in-positive-integ%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
$endgroup$
add a comment |
1
$begingroup$
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
$endgroup$
add a comment |
1
1
1
$begingroup$
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
$endgroup$
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
answered Jan 5 at 18:25
EuxhenHEuxhenH
482210
482210
add a comment |
add a comment |
1
$begingroup$
It seems that $frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :
(1,1,1,1)
(2,2,2,2)
(3,2,2,4)
(3,3,3,3)
(5,3,5,6)
(8,3,9,6)
(4,4,4,4)
(6,4,4,8)
(5,5,3,6)
(5,5,5,5)
(6,6,6,6)
(7,7,7,7)
(8,7,9,4)
(8,8,8,8)
(8,9,3,6)
(8,9,7,4)
(9,9,9,9)
answered Jan 5 at 18:22
YoshiYoshi
133
$endgroup$
add a comment |
1
$begingroup$
It seems that $frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :
(1,1,1,1)
(2,2,2,2)
(3,2,2,4)
(3,3,3,3)
(5,3,5,6)
(8,3,9,6)
(4,4,4,4)
(6,4,4,8)
(5,5,3,6)
(5,5,5,5)
(6,6,6,6)
(7,7,7,7)
(8,7,9,4)
(8,8,8,8)
(8,9,3,6)
(8,9,7,4)
(9,9,9,9)
answered Jan 5 at 18:22
YoshiYoshi
133
$endgroup$
add a comment |
1
1
1
$begingroup$
It seems that $frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :
(1,1,1,1)
(2,2,2,2)
(3,2,2,4)
(3,3,3,3)
(5,3,5,6)
(8,3,9,6)
(4,4,4,4)
(6,4,4,8)
(5,5,3,6)
(5,5,5,5)
(6,6,6,6)
(7,7,7,7)
(8,7,9,4)
(8,8,8,8)
(8,9,3,6)
(8,9,7,4)
(9,9,9,9)
answered Jan 5 at 18:22
YoshiYoshi
133
$endgroup$
It seems that $frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :
(1,1,1,1)
(2,2,2,2)
(3,2,2,4)
(3,3,3,3)
(5,3,5,6)
(8,3,9,6)
(4,4,4,4)
(6,4,4,8)
(5,5,3,6)
(5,5,5,5)
(6,6,6,6)
(7,7,7,7)
(8,7,9,4)
(8,8,8,8)
(8,9,3,6)
(8,9,7,4)
(9,9,9,9)
answered Jan 5 at 18:22
YoshiYoshi
133
edited Jan 5 at 18:31
answered Jan 5 at 18:22
YoshiYoshi
133
answered Jan 5 at 18:22
YoshiYoshi
133
answered Jan 5 at 18:22
YoshiYoshi
133
133
add a comment |
add a comment |
0
$begingroup$
I get it now. Begin with any triple such that $gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy.
$$ k = frac{2b^2+2c^2+d^2}{gcd left( 2b^2+2c^2+d^2 ;, ;2b^3+2c^3+d^3 right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
0
$begingroup$
I get it now. Begin with any triple such that $gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy.
$$ k = frac{2b^2+2c^2+d^2}{gcd left( 2b^2+2c^2+d^2 ;, ;2b^3+2c^3+d^3 right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
0
0
0
$begingroup$
I get it now. Begin with any triple such that $gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy.
$$ k = frac{2b^2+2c^2+d^2}{gcd left( 2b^2+2c^2+d^2 ;, ;2b^3+2c^3+d^3 right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
$endgroup$
I get it now. Begin with any triple such that $gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy.
$$ k = frac{2b^2+2c^2+d^2}{gcd left( 2b^2+2c^2+d^2 ;, ;2b^3+2c^3+d^3 right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
edited Jan 5 at 20:35
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
answered Jan 5 at 19:12
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
0
$begingroup$
For the equation.
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
You can record this parameterization.
$$a=p^3-3s^3$$
$$b=p^3+sp^2-9s^3$$
$$c=p(p^2-6s^2)$$
$$d=p^3-2sp^2+9s^3$$
answered Jan 6 at 9:03
individindivid
3,2621916
$endgroup$
add a comment |
0
$begingroup$
For the equation.
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
You can record this parameterization.
$$a=p^3-3s^3$$
$$b=p^3+sp^2-9s^3$$
$$c=p(p^2-6s^2)$$
$$d=p^3-2sp^2+9s^3$$
answered Jan 6 at 9:03
individindivid
3,2621916
$endgroup$
add a comment |
0
0
0
$begingroup$
For the equation.
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
You can record this parameterization.
$$a=p^3-3s^3$$
$$b=p^3+sp^2-9s^3$$
$$c=p(p^2-6s^2)$$
$$d=p^3-2sp^2+9s^3$$
answered Jan 6 at 9:03
individindivid
3,2621916
$endgroup$
For the equation.
$$a=frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
You can record this parameterization.
$$a=p^3-3s^3$$
$$b=p^3+sp^2-9s^3$$
$$c=p(p^2-6s^2)$$
$$d=p^3-2sp^2+9s^3$$
answered Jan 6 at 9:03
individindivid
3,2621916
answered Jan 6 at 9:03
individindivid
3,2621916
answered Jan 6 at 9:03
individindivid
3,2621916
answered Jan 6 at 9:03
individindivid
3,2621916
3,2621916
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062994%2fare-there-solutions-to-a-frac2b32c3d32b22c2d2-in-positive-integ%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@Peter can these solutions be found only by the brute force of a computer?
$endgroup$
– user631773
Jan 5 at 18:08
$begingroup$
$(2, 2, 2, 2)$ is also a solution where $d>1$.
$endgroup$
– EuxhenH
Jan 5 at 18:23
$begingroup$
(5,4,6,1) is not a solution
$endgroup$
– Will Jagy
Jan 5 at 18:33