How do I prove the following using Fermat's theorem on sums of two squares?
For naturals $x_i,,y_i$ with $1le ile 2019$, $prod_i (2x_i^2+3y_i^2)$ is not a perfect square.
number-theory
edited Dec 9 at 11:16
J.G.
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
add a comment |
For naturals $x_i,,y_i$ with $1le ile 2019$, $prod_i (2x_i^2+3y_i^2)$ is not a perfect square.
number-theory
edited Dec 9 at 11:16
J.G.
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
add a comment |
For naturals $x_i,,y_i$ with $1le ile 2019$, $prod_i (2x_i^2+3y_i^2)$ is not a perfect square.
number-theory
edited Dec 9 at 11:16
J.G.
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
For naturals $x_i,,y_i$ with $1le ile 2019$, $prod_i (2x_i^2+3y_i^2)$ is not a perfect square.
number-theory
number-theory
edited Dec 9 at 11:16
J.G.
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
edited Dec 9 at 11:16
J.G.
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
edited Dec 9 at 11:16
J.G.
22.4k22035
edited Dec 9 at 11:16
J.G.
22.4k22035
edited Dec 9 at 11:16
J.G.
22.4k22035
22.4k22035
asked Dec 9 at 11:10
Piotr Wolski
41
asked Dec 9 at 11:10
Piotr Wolski
41
asked Dec 9 at 11:10
Piotr Wolski
41
41
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
you might as well assume that each pair $(x_i, y_i)$ is coprime, otherwise you just divide out by a long string of squares. Then your product is not divisible by any prime $r equiv 13, 17,19,23 pmod {24}.$ There is no restriction on prime factors $q equiv 1,7 pmod {24}.$
However, if we let $M$ be the set of primes represented by $2 x^2 + 3 y^2,$ we see that $M$ consists of $2,3$ themselves, then all $p equiv 5, 11 pmod{24}.$
The rule for numbers represented as $2u^2 + 3 v^2$ is that the sum of the exponents of primes from $M$ must be odd. That is for a single number. You have an odd count of those, so it is still true that the sum of exponents of primes from $M$ in the prime factorization of your big number must be odd. In particular, there is then at least one such prime factor with odd exponent, and your big product is not a square.
The basic theory here is treated, for example, in Cox, Primes of the Form $x^2 + n y^2$
Side note. A number $x^2 + 6 y^2$ with $gcd(x,y) = 1$ has the sum of $M$ exponents even.
Primes $2 u^2 + 3 v^2$
2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197, 227, 251, 269, 293, 317,
347, 389, 419, 443, 461, 467, 491, 509, 557, 563,
587, 653, 659, 677, 683, 701, 773, 797, 821, 827,
941, 947, 971, 1013, 1019, 1061, 1091, 1109, 1163, 1181,
=======================================================================
numbers $2 x^2 + 3 y^2$ with $gcd(x,y) = 1$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
2 0 3
Discriminant 24
Maximum number represented?
1000
2 = 2
3 = 3
5 = 5
11 = 11
14 = 2 * 7
21 = 3 * 7
29 = 29
30 = 2 * 3 * 5
35 = 5 * 7
50 = 2 * 5^2
53 = 53
59 = 59
62 = 2 * 31
66 = 2 * 3 * 11
75 = 3 * 5^2
77 = 7 * 11
83 = 83
93 = 3 * 31
98 = 2 * 7^2
101 = 101
107 = 107
110 = 2 * 5 * 11
125 = 5^3
131 = 131
146 = 2 * 73
147 = 3 * 7^2
149 = 149
155 = 5 * 31
158 = 2 * 79
165 = 3 * 5 * 11
173 = 173
174 = 2 * 3 * 29
179 = 179
194 = 2 * 97
197 = 197
203 = 7 * 29
206 = 2 * 103
210 = 2 * 3 * 5 * 7
219 = 3 * 73
227 = 227
237 = 3 * 79
242 = 2 * 11^2
245 = 5 * 7^2
251 = 251
254 = 2 * 127
269 = 269
275 = 5^2 * 11
290 = 2 * 5 * 29
291 = 3 * 97
293 = 293
302 = 2 * 151
309 = 3 * 103
317 = 317
318 = 2 * 3 * 53
341 = 11 * 31
347 = 347
350 = 2 * 5^2 * 7
354 = 2 * 3 * 59
363 = 3 * 11^2
365 = 5 * 73
371 = 7 * 53
381 = 3 * 127
386 = 2 * 193
389 = 389
395 = 5 * 79
398 = 2 * 199
413 = 7 * 59
419 = 419
434 = 2 * 7 * 31
435 = 3 * 5 * 29
443 = 443
446 = 2 * 223
453 = 3 * 151
461 = 461
462 = 2 * 3 * 7 * 11
467 = 467
482 = 2 * 241
485 = 5 * 97
491 = 491
498 = 2 * 3 * 83
509 = 509
515 = 5 * 103
525 = 3 * 5^2 * 7
530 = 2 * 5 * 53
539 = 7^2 * 11
542 = 2 * 271
557 = 557
563 = 563
579 = 3 * 193
581 = 7 * 83
587 = 587
590 = 2 * 5 * 59
597 = 3 * 199
605 = 5 * 11^2
606 = 2 * 3 * 101
626 = 2 * 313
635 = 5 * 127
638 = 2 * 11 * 29
642 = 2 * 3 * 107
651 = 3 * 7 * 31
653 = 653
659 = 659
669 = 3 * 223
674 = 2 * 337
677 = 677
683 = 683
686 = 2 * 7^3
701 = 701
707 = 7 * 101
723 = 3 * 241
725 = 5^2 * 29
734 = 2 * 367
749 = 7 * 107
750 = 2 * 3 * 5^3
755 = 5 * 151
770 = 2 * 5 * 7 * 11
773 = 773
786 = 2 * 3 * 131
795 = 3 * 5 * 53
797 = 797
803 = 11 * 73
813 = 3 * 271
818 = 2 * 409
821 = 821
827 = 827
830 = 2 * 5 * 83
866 = 2 * 433
869 = 11 * 79
875 = 5^3 * 7
878 = 2 * 439
885 = 3 * 5 * 59
894 = 2 * 3 * 149
899 = 29 * 31
914 = 2 * 457
917 = 7 * 131
926 = 2 * 463
930 = 2 * 3 * 5 * 31
939 = 3 * 313
941 = 941
947 = 947
957 = 3 * 11 * 29
965 = 5 * 193
971 = 971
974 = 2 * 487
995 = 5 * 199
======================================================================
answered Dec 9 at 16:17
Will Jagy
101k599199
add a comment |
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1 Answer
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you might as well assume that each pair $(x_i, y_i)$ is coprime, otherwise you just divide out by a long string of squares. Then your product is not divisible by any prime $r equiv 13, 17,19,23 pmod {24}.$ There is no restriction on prime factors $q equiv 1,7 pmod {24}.$
However, if we let $M$ be the set of primes represented by $2 x^2 + 3 y^2,$ we see that $M$ consists of $2,3$ themselves, then all $p equiv 5, 11 pmod{24}.$
The rule for numbers represented as $2u^2 + 3 v^2$ is that the sum of the exponents of primes from $M$ must be odd. That is for a single number. You have an odd count of those, so it is still true that the sum of exponents of primes from $M$ in the prime factorization of your big number must be odd. In particular, there is then at least one such prime factor with odd exponent, and your big product is not a square.
The basic theory here is treated, for example, in Cox, Primes of the Form $x^2 + n y^2$
Side note. A number $x^2 + 6 y^2$ with $gcd(x,y) = 1$ has the sum of $M$ exponents even.
Primes $2 u^2 + 3 v^2$
2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197, 227, 251, 269, 293, 317,
347, 389, 419, 443, 461, 467, 491, 509, 557, 563,
587, 653, 659, 677, 683, 701, 773, 797, 821, 827,
941, 947, 971, 1013, 1019, 1061, 1091, 1109, 1163, 1181,
=======================================================================
numbers $2 x^2 + 3 y^2$ with $gcd(x,y) = 1$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
2 0 3
Discriminant 24
Maximum number represented?
1000
2 = 2
3 = 3
5 = 5
11 = 11
14 = 2 * 7
21 = 3 * 7
29 = 29
30 = 2 * 3 * 5
35 = 5 * 7
50 = 2 * 5^2
53 = 53
59 = 59
62 = 2 * 31
66 = 2 * 3 * 11
75 = 3 * 5^2
77 = 7 * 11
83 = 83
93 = 3 * 31
98 = 2 * 7^2
101 = 101
107 = 107
110 = 2 * 5 * 11
125 = 5^3
131 = 131
146 = 2 * 73
147 = 3 * 7^2
149 = 149
155 = 5 * 31
158 = 2 * 79
165 = 3 * 5 * 11
173 = 173
174 = 2 * 3 * 29
179 = 179
194 = 2 * 97
197 = 197
203 = 7 * 29
206 = 2 * 103
210 = 2 * 3 * 5 * 7
219 = 3 * 73
227 = 227
237 = 3 * 79
242 = 2 * 11^2
245 = 5 * 7^2
251 = 251
254 = 2 * 127
269 = 269
275 = 5^2 * 11
290 = 2 * 5 * 29
291 = 3 * 97
293 = 293
302 = 2 * 151
309 = 3 * 103
317 = 317
318 = 2 * 3 * 53
341 = 11 * 31
347 = 347
350 = 2 * 5^2 * 7
354 = 2 * 3 * 59
363 = 3 * 11^2
365 = 5 * 73
371 = 7 * 53
381 = 3 * 127
386 = 2 * 193
389 = 389
395 = 5 * 79
398 = 2 * 199
413 = 7 * 59
419 = 419
434 = 2 * 7 * 31
435 = 3 * 5 * 29
443 = 443
446 = 2 * 223
453 = 3 * 151
461 = 461
462 = 2 * 3 * 7 * 11
467 = 467
482 = 2 * 241
485 = 5 * 97
491 = 491
498 = 2 * 3 * 83
509 = 509
515 = 5 * 103
525 = 3 * 5^2 * 7
530 = 2 * 5 * 53
539 = 7^2 * 11
542 = 2 * 271
557 = 557
563 = 563
579 = 3 * 193
581 = 7 * 83
587 = 587
590 = 2 * 5 * 59
597 = 3 * 199
605 = 5 * 11^2
606 = 2 * 3 * 101
626 = 2 * 313
635 = 5 * 127
638 = 2 * 11 * 29
642 = 2 * 3 * 107
651 = 3 * 7 * 31
653 = 653
659 = 659
669 = 3 * 223
674 = 2 * 337
677 = 677
683 = 683
686 = 2 * 7^3
701 = 701
707 = 7 * 101
723 = 3 * 241
725 = 5^2 * 29
734 = 2 * 367
749 = 7 * 107
750 = 2 * 3 * 5^3
755 = 5 * 151
770 = 2 * 5 * 7 * 11
773 = 773
786 = 2 * 3 * 131
795 = 3 * 5 * 53
797 = 797
803 = 11 * 73
813 = 3 * 271
818 = 2 * 409
821 = 821
827 = 827
830 = 2 * 5 * 83
866 = 2 * 433
869 = 11 * 79
875 = 5^3 * 7
878 = 2 * 439
885 = 3 * 5 * 59
894 = 2 * 3 * 149
899 = 29 * 31
914 = 2 * 457
917 = 7 * 131
926 = 2 * 463
930 = 2 * 3 * 5 * 31
939 = 3 * 313
941 = 941
947 = 947
957 = 3 * 11 * 29
965 = 5 * 193
971 = 971
974 = 2 * 487
995 = 5 * 199
======================================================================
answered Dec 9 at 16:17
Will Jagy
101k599199
add a comment |
you might as well assume that each pair $(x_i, y_i)$ is coprime, otherwise you just divide out by a long string of squares. Then your product is not divisible by any prime $r equiv 13, 17,19,23 pmod {24}.$ There is no restriction on prime factors $q equiv 1,7 pmod {24}.$
However, if we let $M$ be the set of primes represented by $2 x^2 + 3 y^2,$ we see that $M$ consists of $2,3$ themselves, then all $p equiv 5, 11 pmod{24}.$
The rule for numbers represented as $2u^2 + 3 v^2$ is that the sum of the exponents of primes from $M$ must be odd. That is for a single number. You have an odd count of those, so it is still true that the sum of exponents of primes from $M$ in the prime factorization of your big number must be odd. In particular, there is then at least one such prime factor with odd exponent, and your big product is not a square.
The basic theory here is treated, for example, in Cox, Primes of the Form $x^2 + n y^2$
Side note. A number $x^2 + 6 y^2$ with $gcd(x,y) = 1$ has the sum of $M$ exponents even.
Primes $2 u^2 + 3 v^2$
2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197, 227, 251, 269, 293, 317,
347, 389, 419, 443, 461, 467, 491, 509, 557, 563,
587, 653, 659, 677, 683, 701, 773, 797, 821, 827,
941, 947, 971, 1013, 1019, 1061, 1091, 1109, 1163, 1181,
=======================================================================
numbers $2 x^2 + 3 y^2$ with $gcd(x,y) = 1$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
2 0 3
Discriminant 24
Maximum number represented?
1000
2 = 2
3 = 3
5 = 5
11 = 11
14 = 2 * 7
21 = 3 * 7
29 = 29
30 = 2 * 3 * 5
35 = 5 * 7
50 = 2 * 5^2
53 = 53
59 = 59
62 = 2 * 31
66 = 2 * 3 * 11
75 = 3 * 5^2
77 = 7 * 11
83 = 83
93 = 3 * 31
98 = 2 * 7^2
101 = 101
107 = 107
110 = 2 * 5 * 11
125 = 5^3
131 = 131
146 = 2 * 73
147 = 3 * 7^2
149 = 149
155 = 5 * 31
158 = 2 * 79
165 = 3 * 5 * 11
173 = 173
174 = 2 * 3 * 29
179 = 179
194 = 2 * 97
197 = 197
203 = 7 * 29
206 = 2 * 103
210 = 2 * 3 * 5 * 7
219 = 3 * 73
227 = 227
237 = 3 * 79
242 = 2 * 11^2
245 = 5 * 7^2
251 = 251
254 = 2 * 127
269 = 269
275 = 5^2 * 11
290 = 2 * 5 * 29
291 = 3 * 97
293 = 293
302 = 2 * 151
309 = 3 * 103
317 = 317
318 = 2 * 3 * 53
341 = 11 * 31
347 = 347
350 = 2 * 5^2 * 7
354 = 2 * 3 * 59
363 = 3 * 11^2
365 = 5 * 73
371 = 7 * 53
381 = 3 * 127
386 = 2 * 193
389 = 389
395 = 5 * 79
398 = 2 * 199
413 = 7 * 59
419 = 419
434 = 2 * 7 * 31
435 = 3 * 5 * 29
443 = 443
446 = 2 * 223
453 = 3 * 151
461 = 461
462 = 2 * 3 * 7 * 11
467 = 467
482 = 2 * 241
485 = 5 * 97
491 = 491
498 = 2 * 3 * 83
509 = 509
515 = 5 * 103
525 = 3 * 5^2 * 7
530 = 2 * 5 * 53
539 = 7^2 * 11
542 = 2 * 271
557 = 557
563 = 563
579 = 3 * 193
581 = 7 * 83
587 = 587
590 = 2 * 5 * 59
597 = 3 * 199
605 = 5 * 11^2
606 = 2 * 3 * 101
626 = 2 * 313
635 = 5 * 127
638 = 2 * 11 * 29
642 = 2 * 3 * 107
651 = 3 * 7 * 31
653 = 653
659 = 659
669 = 3 * 223
674 = 2 * 337
677 = 677
683 = 683
686 = 2 * 7^3
701 = 701
707 = 7 * 101
723 = 3 * 241
725 = 5^2 * 29
734 = 2 * 367
749 = 7 * 107
750 = 2 * 3 * 5^3
755 = 5 * 151
770 = 2 * 5 * 7 * 11
773 = 773
786 = 2 * 3 * 131
795 = 3 * 5 * 53
797 = 797
803 = 11 * 73
813 = 3 * 271
818 = 2 * 409
821 = 821
827 = 827
830 = 2 * 5 * 83
866 = 2 * 433
869 = 11 * 79
875 = 5^3 * 7
878 = 2 * 439
885 = 3 * 5 * 59
894 = 2 * 3 * 149
899 = 29 * 31
914 = 2 * 457
917 = 7 * 131
926 = 2 * 463
930 = 2 * 3 * 5 * 31
939 = 3 * 313
941 = 941
947 = 947
957 = 3 * 11 * 29
965 = 5 * 193
971 = 971
974 = 2 * 487
995 = 5 * 199
======================================================================
answered Dec 9 at 16:17
Will Jagy
101k599199
add a comment |
you might as well assume that each pair $(x_i, y_i)$ is coprime, otherwise you just divide out by a long string of squares. Then your product is not divisible by any prime $r equiv 13, 17,19,23 pmod {24}.$ There is no restriction on prime factors $q equiv 1,7 pmod {24}.$
However, if we let $M$ be the set of primes represented by $2 x^2 + 3 y^2,$ we see that $M$ consists of $2,3$ themselves, then all $p equiv 5, 11 pmod{24}.$
The rule for numbers represented as $2u^2 + 3 v^2$ is that the sum of the exponents of primes from $M$ must be odd. That is for a single number. You have an odd count of those, so it is still true that the sum of exponents of primes from $M$ in the prime factorization of your big number must be odd. In particular, there is then at least one such prime factor with odd exponent, and your big product is not a square.
The basic theory here is treated, for example, in Cox, Primes of the Form $x^2 + n y^2$
Side note. A number $x^2 + 6 y^2$ with $gcd(x,y) = 1$ has the sum of $M$ exponents even.
Primes $2 u^2 + 3 v^2$
2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197, 227, 251, 269, 293, 317,
347, 389, 419, 443, 461, 467, 491, 509, 557, 563,
587, 653, 659, 677, 683, 701, 773, 797, 821, 827,
941, 947, 971, 1013, 1019, 1061, 1091, 1109, 1163, 1181,
=======================================================================
numbers $2 x^2 + 3 y^2$ with $gcd(x,y) = 1$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
2 0 3
Discriminant 24
Maximum number represented?
1000
2 = 2
3 = 3
5 = 5
11 = 11
14 = 2 * 7
21 = 3 * 7
29 = 29
30 = 2 * 3 * 5
35 = 5 * 7
50 = 2 * 5^2
53 = 53
59 = 59
62 = 2 * 31
66 = 2 * 3 * 11
75 = 3 * 5^2
77 = 7 * 11
83 = 83
93 = 3 * 31
98 = 2 * 7^2
101 = 101
107 = 107
110 = 2 * 5 * 11
125 = 5^3
131 = 131
146 = 2 * 73
147 = 3 * 7^2
149 = 149
155 = 5 * 31
158 = 2 * 79
165 = 3 * 5 * 11
173 = 173
174 = 2 * 3 * 29
179 = 179
194 = 2 * 97
197 = 197
203 = 7 * 29
206 = 2 * 103
210 = 2 * 3 * 5 * 7
219 = 3 * 73
227 = 227
237 = 3 * 79
242 = 2 * 11^2
245 = 5 * 7^2
251 = 251
254 = 2 * 127
269 = 269
275 = 5^2 * 11
290 = 2 * 5 * 29
291 = 3 * 97
293 = 293
302 = 2 * 151
309 = 3 * 103
317 = 317
318 = 2 * 3 * 53
341 = 11 * 31
347 = 347
350 = 2 * 5^2 * 7
354 = 2 * 3 * 59
363 = 3 * 11^2
365 = 5 * 73
371 = 7 * 53
381 = 3 * 127
386 = 2 * 193
389 = 389
395 = 5 * 79
398 = 2 * 199
413 = 7 * 59
419 = 419
434 = 2 * 7 * 31
435 = 3 * 5 * 29
443 = 443
446 = 2 * 223
453 = 3 * 151
461 = 461
462 = 2 * 3 * 7 * 11
467 = 467
482 = 2 * 241
485 = 5 * 97
491 = 491
498 = 2 * 3 * 83
509 = 509
515 = 5 * 103
525 = 3 * 5^2 * 7
530 = 2 * 5 * 53
539 = 7^2 * 11
542 = 2 * 271
557 = 557
563 = 563
579 = 3 * 193
581 = 7 * 83
587 = 587
590 = 2 * 5 * 59
597 = 3 * 199
605 = 5 * 11^2
606 = 2 * 3 * 101
626 = 2 * 313
635 = 5 * 127
638 = 2 * 11 * 29
642 = 2 * 3 * 107
651 = 3 * 7 * 31
653 = 653
659 = 659
669 = 3 * 223
674 = 2 * 337
677 = 677
683 = 683
686 = 2 * 7^3
701 = 701
707 = 7 * 101
723 = 3 * 241
725 = 5^2 * 29
734 = 2 * 367
749 = 7 * 107
750 = 2 * 3 * 5^3
755 = 5 * 151
770 = 2 * 5 * 7 * 11
773 = 773
786 = 2 * 3 * 131
795 = 3 * 5 * 53
797 = 797
803 = 11 * 73
813 = 3 * 271
818 = 2 * 409
821 = 821
827 = 827
830 = 2 * 5 * 83
866 = 2 * 433
869 = 11 * 79
875 = 5^3 * 7
878 = 2 * 439
885 = 3 * 5 * 59
894 = 2 * 3 * 149
899 = 29 * 31
914 = 2 * 457
917 = 7 * 131
926 = 2 * 463
930 = 2 * 3 * 5 * 31
939 = 3 * 313
941 = 941
947 = 947
957 = 3 * 11 * 29
965 = 5 * 193
971 = 971
974 = 2 * 487
995 = 5 * 199
======================================================================
answered Dec 9 at 16:17
Will Jagy
101k599199
you might as well assume that each pair $(x_i, y_i)$ is coprime, otherwise you just divide out by a long string of squares. Then your product is not divisible by any prime $r equiv 13, 17,19,23 pmod {24}.$ There is no restriction on prime factors $q equiv 1,7 pmod {24}.$
However, if we let $M$ be the set of primes represented by $2 x^2 + 3 y^2,$ we see that $M$ consists of $2,3$ themselves, then all $p equiv 5, 11 pmod{24}.$
The rule for numbers represented as $2u^2 + 3 v^2$ is that the sum of the exponents of primes from $M$ must be odd. That is for a single number. You have an odd count of those, so it is still true that the sum of exponents of primes from $M$ in the prime factorization of your big number must be odd. In particular, there is then at least one such prime factor with odd exponent, and your big product is not a square.
The basic theory here is treated, for example, in Cox, Primes of the Form $x^2 + n y^2$
Side note. A number $x^2 + 6 y^2$ with $gcd(x,y) = 1$ has the sum of $M$ exponents even.
Primes $2 u^2 + 3 v^2$
2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197, 227, 251, 269, 293, 317,
347, 389, 419, 443, 461, 467, 491, 509, 557, 563,
587, 653, 659, 677, 683, 701, 773, 797, 821, 827,
941, 947, 971, 1013, 1019, 1061, 1091, 1109, 1163, 1181,
=======================================================================
numbers $2 x^2 + 3 y^2$ with $gcd(x,y) = 1$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
2 0 3
Discriminant 24
Maximum number represented?
1000
2 = 2
3 = 3
5 = 5
11 = 11
14 = 2 * 7
21 = 3 * 7
29 = 29
30 = 2 * 3 * 5
35 = 5 * 7
50 = 2 * 5^2
53 = 53
59 = 59
62 = 2 * 31
66 = 2 * 3 * 11
75 = 3 * 5^2
77 = 7 * 11
83 = 83
93 = 3 * 31
98 = 2 * 7^2
101 = 101
107 = 107
110 = 2 * 5 * 11
125 = 5^3
131 = 131
146 = 2 * 73
147 = 3 * 7^2
149 = 149
155 = 5 * 31
158 = 2 * 79
165 = 3 * 5 * 11
173 = 173
174 = 2 * 3 * 29
179 = 179
194 = 2 * 97
197 = 197
203 = 7 * 29
206 = 2 * 103
210 = 2 * 3 * 5 * 7
219 = 3 * 73
227 = 227
237 = 3 * 79
242 = 2 * 11^2
245 = 5 * 7^2
251 = 251
254 = 2 * 127
269 = 269
275 = 5^2 * 11
290 = 2 * 5 * 29
291 = 3 * 97
293 = 293
302 = 2 * 151
309 = 3 * 103
317 = 317
318 = 2 * 3 * 53
341 = 11 * 31
347 = 347
350 = 2 * 5^2 * 7
354 = 2 * 3 * 59
363 = 3 * 11^2
365 = 5 * 73
371 = 7 * 53
381 = 3 * 127
386 = 2 * 193
389 = 389
395 = 5 * 79
398 = 2 * 199
413 = 7 * 59
419 = 419
434 = 2 * 7 * 31
435 = 3 * 5 * 29
443 = 443
446 = 2 * 223
453 = 3 * 151
461 = 461
462 = 2 * 3 * 7 * 11
467 = 467
482 = 2 * 241
485 = 5 * 97
491 = 491
498 = 2 * 3 * 83
509 = 509
515 = 5 * 103
525 = 3 * 5^2 * 7
530 = 2 * 5 * 53
539 = 7^2 * 11
542 = 2 * 271
557 = 557
563 = 563
579 = 3 * 193
581 = 7 * 83
587 = 587
590 = 2 * 5 * 59
597 = 3 * 199
605 = 5 * 11^2
606 = 2 * 3 * 101
626 = 2 * 313
635 = 5 * 127
638 = 2 * 11 * 29
642 = 2 * 3 * 107
651 = 3 * 7 * 31
653 = 653
659 = 659
669 = 3 * 223
674 = 2 * 337
677 = 677
683 = 683
686 = 2 * 7^3
701 = 701
707 = 7 * 101
723 = 3 * 241
725 = 5^2 * 29
734 = 2 * 367
749 = 7 * 107
750 = 2 * 3 * 5^3
755 = 5 * 151
770 = 2 * 5 * 7 * 11
773 = 773
786 = 2 * 3 * 131
795 = 3 * 5 * 53
797 = 797
803 = 11 * 73
813 = 3 * 271
818 = 2 * 409
821 = 821
827 = 827
830 = 2 * 5 * 83
866 = 2 * 433
869 = 11 * 79
875 = 5^3 * 7
878 = 2 * 439
885 = 3 * 5 * 59
894 = 2 * 3 * 149
899 = 29 * 31
914 = 2 * 457
917 = 7 * 131
926 = 2 * 463
930 = 2 * 3 * 5 * 31
939 = 3 * 313
941 = 941
947 = 947
957 = 3 * 11 * 29
965 = 5 * 193
971 = 971
974 = 2 * 487
995 = 5 * 199
======================================================================
answered Dec 9 at 16:17
Will Jagy
101k599199
edited Dec 9 at 16:26
answered Dec 9 at 16:17
Will Jagy
101k599199
answered Dec 9 at 16:17
Will Jagy
101k599199
answered Dec 9 at 16:17
Will Jagy
101k599199
101k599199
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