Prove that if $\gcd(a,b) = 1$, then $\gcd(ab,a^2+b^2)=1.$

Question

I was reading a Proof that stated

Since $\gcd(a, b) = 1$, it follows that $\gcd(ab, a^2 + b^2) = 1$.

I can't figure out why this statement is true. I tried to factor $ab$ and $a^2+b^2$, but I don't think that you can really reduce $a^2$ and $b^2$ because it's just addition. I'm not really sure how to go about this. Any help?

Answer 1

I always prefer Bézout's Identity proofs.

Lemma 1: If $\gcd(c,d)=1$ then $\gcd(c,ck+d)=1$ for any $k$.

Proof: If $cx+dy=1$, then $c(x-ky)+(ck+d)y=1$.

Lemma 2: If $\gcd(c_1,d)=1$ and $\gcd(c_2,d)=1$ then $\gcd(c_1c_2,d)=1$.

Proof: Solve $c_1x_1+dy_1=1$ and $c_2x_2+dy_2=1$. Multiplying, and you get:

$$c_1c_2(x_1y_1)+d(y_2c_1x_1+y_1c_2x_2+dy_1y_2)=1$$

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

Proof: Lemma 2 implies that $\gcd(a,b^2)=1$. Lemma 1 implies that $\gcd(a,a^2+b^2)=1$.

Likewise, we get that $\gcd(b,a^2+b^2)=1$.

Then, by Lemma 2, we have that $\gcd(ab,a^2+b^2)=1$.

Answer 2

Let $d=gcd(ab,a^2+b^2)$. Then $d\mid ab$ and $d\mid a^2+b^2$. For a prime divisor $p\mid d$ with $p\mid ab$ we have $p\mid a$ or $p\mid b$. Suppose that $p\mid a$. Then $p\mid a^2$, and because $p\mid a^2+b^2$ we also get $p\mid b^2$. Similarly $p\mid b$ implies that $p\mid a^2$. Hence we obtain $$ p\mid gcd(a^2,b^2)=gcd(a,b)=1. $$ For the last step, see this duplicate: Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.

Answer 3

Remark: every natural number greater than $1$, have a prime factor.

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Euclid's Lemma: Let $a$ and $b$ to be integers and let $p$ to be a prime number.

If $p \mid ab$ then $p \mid a$ or $p \mid b$.

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Lemma(2): Let $b$ to be an integer and let $p$ to be a prime number.

If $p \mid b^2$ then $p \mid b$.

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Suppose on contrary that $\gcd(ab, a^2+b^2) \geq 1$, so by the above remark it must have a prime factor $p$.

Notice that $p \mid ab$, without loss of generality by the Euclid's lemma we can assume that $p \mid a$, so we have: $p \mid a^2$.

On the otherhand $p \mid a^2+b^2$, so we have $p \mid (a^2+b^2)-a^2$, i.e. $p \mid b^2$, and by the lemma(2) then $p \mid b$.

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So we have: $p \mid a$ and $p \mid b$; but notice that, it is an obvious contradiction with the assumption that $\gcd(a,b)=1$.

Answer 4

Set $d:=gcd(ab,a^2+b^2)$ and assume $d\neq 1.$ This means that there exists a prime $p$ with \begin{equation}p|d|ab, \end{equation}which implies $p|a$ or $p|b.$ Wlog we can say $p|a.$ Using this and \begin{equation}p|d|a^2+b^2 \end{equation}implies $p|b^2$ and therefore $p|b$, because p is prime. But this means that $p$ is a number at least $2$ and divides $a,b.$ But this is a contradiction to $gcd(a,b)=1.$

Answer 5

If $p$ is a prime and $p|ab, p|(a^2+b^2)$ then $p|(a^2+2ab+b^2)$, so $p|a+b$.

But then $p|ab+b^2$ and $p|ab+a^2$, so $p|b^2$, $p|a^2$, implying $p|MCD(a^2,b^2)=1$, contradiction.

Then $p$ doesn't exist, and the problem is solved!