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If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.

So far, I let

$d=(a^2+b^2,a+b)$

$\implies d|(a^2+b^2-(a+b)^2)$

$\implies d|(a^2+b^2-(a^2+2ab+b^2))$

$\implies d|(-2ab)$

I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Alternatively, is there some other way of showing it?

EDIT(10:41PM):

Just noticed that $(a^2+b^2\pm (a+b)(a-b),a+b)$ results in $d|2a^2$ and $d|2b^2$.

$\implies d|(2a^2,2b^2)$.

$\implies d|2(a^2,b^2)$.

Now I just have to fully understand why

$(a,b)=1\implies (a^2,b^2)=1$.

I sort of intuitively understand this.

Martin Sleziak
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4 Answers4

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Hint: What contradiction can you get if $d|a$ or $d|b$ but $d \neq1$?

2

Lemma 1: If $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.

Proof: We will prove the contrapositive. Suppose $\gcd(a^2,b^2)>1$. Then $a^2$ and $b^2$ must have a prime common divisor, say $p$. Since $p|a^2$, Euclid's lemma implies that $p|a$. Similarly, $p|b$. Thus $p$ is a common divisor of $a,b$ and so $\gcd(a,b)>1$.

Lemma 2: If $\gcd(u,v)=1$, then $\gcd(u+v,u-v) \leq 2$.

Proof: Let $d$ be a common divisor of $u+v$ and $u-v$. Then $d|(u+v+u-v)$ and so $d|2u$. Moreover, $d|(u+v-(u-v))$ and so $d|2v$. Thus $d|\gcd(2u,2v)$. But $\gcd(2u,2v)=2\gcd(u,v)=2$. It follows that $\gcd(u+v,u-v) \leq 2$.

Given these two lemmas, we'll prove the problem statement. Since $\gcd(a,b)=1$, $\gcd(a^2,b^2)=1$ by Lemma 1. Now, by setting $u=a^2$, $v=b^2$ in Lemma 2, we can see that $\gcd(a^2+b^2,a^2-b^2) \leq 2$.

Finally, suppose $d$ is a common divisor of $a^2+b^2$ and $a+b$. Because $d|(a+b)$, $d|(a+b)(a-b)=a^2-b^2$. So $d$ is a common divisor of $a^2+b^2$ and $a^2-b^2$. It follows that $d \leq 2$, completing the proof.

Micah
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  • I don't quite get the first bullet. The second bullet I agree with. I definitely agree with the third bullet since $(a^2-b^2)=(a+b)(a-b)$. –  Oct 14 '14 at 05:16
  • By "don't get", do you mean you don't understand why it's true, or why it's helpful? – Micah Oct 14 '14 at 05:17
  • Sorry, I meant that I don't understand why it is true. I think it has something to do with the fact that a and b have no common factors besides 1, so when they are squared, they still share 1 as their only common factor. –  Oct 14 '14 at 05:20
  • I also don't see the usefulness. –  Oct 14 '14 at 05:20
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    That's the right idea. To make it rigorous, consider only prime common factors; if $p$ is prime and $p|a^2$, then certainly $p|a$ by Euclid's lemma, and so any prime common divisor of $a^2$ and $b^2$ must also be a common divisor of $a$ and $b$. To see why it's useful, think about what happens when you feed it into the second bullet point... – Micah Oct 14 '14 at 05:35
  • I am not sure what you are getting at with the second bullet, but the first bullet is definitely helping me. I just need to make sure I completely understand the first bullet, and once I do, I will complete the proof. –  Oct 14 '14 at 05:49
  • What I was getting at is that, by the first two bullets, $\gcd(a^2+b^2,a^2-b^2) \leq 2$. Now the third bullet gives you the desired result. I think your second solution is morally equivalent to this, though... – Micah Oct 14 '14 at 05:57
  • I don't understand. From what you've said, I think your logic is something like, $(a+b)|(a^2-b^2)$ and $(a+b,a-b)\le2\implies(a+b)\le2$. I don't follow, but I think you actually mean something that I am just not seeing. –  Oct 15 '14 at 19:47
  • See my updated answer. – Micah Oct 15 '14 at 20:13
  • Ok, I think I can follow that. It is very clear. –  Oct 17 '14 at 16:00
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Consider the case where a and b are both odd. In this case, since d divides (-2ab), and a and b are coprime, the condition becomes d divides a or d divides b or d divides 2. If d divides a, then d | ( (a+b)-a ), and thus d divides b, a contradiction since a and b are coprime. Hence, d divides 2; d is 1 or 2.

Next consider the case where a even, b odd. (The case where b even, a odd is similar.) We express a as (2^n)(c), where c is an odd positive integer and n is positive integer. Then d divides (-2ab) is equivalent to d divides (2a) or d divides b (since 2a and b are coprime). If d divides b, then by similar reasoning as above d would divide a, leading to a contradiction. The remaining case is where d contains 2^(n+1) as a factor; however that cannot be true as d divides (a+b) which is odd. Therefore, as per the first case d divides 2; d is 1 or 2.

Sean Lo
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Many gcd questions can be answered by finding a way to use the Euclidean algorithm. By the Euclidean algorithm, I mean $(a, aq+r) = (a,r)$.

Here, $a$ is played by $a+b$, $q$ is played by $a-b$, and $r$ is played by $2b^2$. The Euclidean algorithm then implies that $$(a+b, (a+b)(a-b) + 2b^2) = (a+b, 2b^2).$$

This reads as $(a^2+b^2, a+b) = (a+b, 2b^2)$. The conclusion should now be clear.

RghtHndSd
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  • It is not clear to me, I'm sorry. So you're saying, in the last equation, that the left and right sides are equal. I think you're meaning that $(a+b,2b^2)$ must be 1 or 2, so it is the same for the left side. Is that right? I am still processing your claims, by the way, so maybe I will understand soon. –  Oct 14 '14 at 04:39
  • @alphagamma: That's correct. Let $p$ be a prime factor of the gcd. Then $p$ divides $2b^2$. If $p$ divides $b^2$, then $p$ divides $b$ and so $p$ does not divide $a+b$. We conclude $p$ does not divide $b^2$, and so it must be that $p$ divides 2. – RghtHndSd Oct 14 '14 at 04:47
  • How can you conclude that $p\not|(a+b)$? I see that if $p|b^2$ then $p|b$, but what if adding $a$ to $b$ makes it so that $p|(a+b)$? –  Oct 14 '14 at 04:59
  • @alphagamma: Then $a$ and $b$ would have a common factor since $p \mid (a + b - b)$. – RghtHndSd Oct 14 '14 at 05:01
  • whoops, forgot that they were relatively prime... (I continue to try and understand) –  Oct 14 '14 at 05:07