proof on difference of two squares and odd integers
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0
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How would you prove that every odd integer is a difference of two squares?
I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.
So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.
proof-writing
asked Dec 4 at 19:27
Carlos Bacca
178116
add a comment |
up vote
0
down vote
favorite
How would you prove that every odd integer is a difference of two squares?
I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.
So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.
proof-writing
asked Dec 4 at 19:27
Carlos Bacca
178116
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How would you prove that every odd integer is a difference of two squares?
I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.
So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.
proof-writing
asked Dec 4 at 19:27
Carlos Bacca
178116
How would you prove that every odd integer is a difference of two squares?
I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.
So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.
proof-writing
proof-writing
asked Dec 4 at 19:27
Carlos Bacca
178116
asked Dec 4 at 19:27
Carlos Bacca
178116
asked Dec 4 at 19:27
Carlos Bacca
178116
asked Dec 4 at 19:27
Carlos Bacca
178116
asked Dec 4 at 19:27
Carlos Bacca
178116
178116
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35
add a comment |
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35
add a comment |
5 Answers
5
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oldest
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0
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accepted
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.
answered Dec 4 at 19:32
KM101
3,812417
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
add a comment |
up vote
0
down vote
Hint:
Compute $(n+1)^2-n^2$.
You should get $2n+1$
answered Dec 4 at 19:29
Mason
1,8881527
add a comment |
up vote
0
down vote
HINT
$$(k+1)^2-k^2=2k+1$$
$$k^2-(k+1)^2=-(2k+1)$$
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
add a comment |
up vote
0
down vote
Hint: What is the difference of squares of two consecutive numbers?
answered Dec 4 at 19:29
Vasily Mitch
1,14837
add a comment |
up vote
0
down vote
$$3=4-1$$
$$2n+1=(n+1)^2-n^2$$
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.
answered Dec 4 at 19:32
KM101
3,812417
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
add a comment |
up vote
0
down vote
accepted
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.
answered Dec 4 at 19:32
KM101
3,812417
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.
answered Dec 4 at 19:32
KM101
3,812417
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.
answered Dec 4 at 19:32
KM101
3,812417
answered Dec 4 at 19:32
KM101
3,812417
answered Dec 4 at 19:32
KM101
3,812417
answered Dec 4 at 19:32
KM101
3,812417
3,812417
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
add a comment |
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
– Carlos Bacca
Dec 4 at 19:38
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
– KM101
Dec 4 at 19:39
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
is that working backwards?
– Carlos Bacca
Dec 4 at 19:40
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
actually ignore me
– Carlos Bacca
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
Nope, it's the continuation of the proof.
– KM101
Dec 4 at 19:41
add a comment |
up vote
0
down vote
Hint:
Compute $(n+1)^2-n^2$.
You should get $2n+1$
answered Dec 4 at 19:29
Mason
1,8881527
add a comment |
up vote
0
down vote
Hint:
Compute $(n+1)^2-n^2$.
You should get $2n+1$
answered Dec 4 at 19:29
Mason
1,8881527
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
Compute $(n+1)^2-n^2$.
You should get $2n+1$
answered Dec 4 at 19:29
Mason
1,8881527
Hint:
Compute $(n+1)^2-n^2$.
You should get $2n+1$
answered Dec 4 at 19:29
Mason
1,8881527
answered Dec 4 at 19:29
Mason
1,8881527
answered Dec 4 at 19:29
Mason
1,8881527
answered Dec 4 at 19:29
Mason
1,8881527
1,8881527
add a comment |
add a comment |
up vote
0
down vote
HINT
$$(k+1)^2-k^2=2k+1$$
$$k^2-(k+1)^2=-(2k+1)$$
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
add a comment |
up vote
0
down vote
HINT
$$(k+1)^2-k^2=2k+1$$
$$k^2-(k+1)^2=-(2k+1)$$
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
$$(k+1)^2-k^2=2k+1$$
$$k^2-(k+1)^2=-(2k+1)$$
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
HINT
$$(k+1)^2-k^2=2k+1$$
$$k^2-(k+1)^2=-(2k+1)$$
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
answered Dec 4 at 19:29
Rhys Hughes
4,6651327
4,6651327
add a comment |
add a comment |
up vote
0
down vote
Hint: What is the difference of squares of two consecutive numbers?
answered Dec 4 at 19:29
Vasily Mitch
1,14837
add a comment |
up vote
0
down vote
Hint: What is the difference of squares of two consecutive numbers?
answered Dec 4 at 19:29
Vasily Mitch
1,14837
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: What is the difference of squares of two consecutive numbers?
answered Dec 4 at 19:29
Vasily Mitch
1,14837
Hint: What is the difference of squares of two consecutive numbers?
answered Dec 4 at 19:29
Vasily Mitch
1,14837
answered Dec 4 at 19:29
Vasily Mitch
1,14837
answered Dec 4 at 19:29
Vasily Mitch
1,14837
answered Dec 4 at 19:29
Vasily Mitch
1,14837
1,14837
add a comment |
add a comment |
up vote
0
down vote
$$3=4-1$$
$$2n+1=(n+1)^2-n^2$$
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
$$3=4-1$$
$$2n+1=(n+1)^2-n^2$$
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
up vote
0
down vote
$$3=4-1$$
$$2n+1=(n+1)^2-n^2$$
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
$$3=4-1$$
$$2n+1=(n+1)^2-n^2$$
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:37
hamam_Abdallah
37.7k21634
37.7k21634
add a comment |
add a comment |
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See this picture: en.wikipedia.org/wiki/Square_number#Properties.
– Michael Hoppe
Dec 4 at 19:35