Prove that $\gcd(a^2, b^2) = \gcd(a, b)^2$

Question

The problem's quite clear. Prove that $$\gcd(a^2, b^2) = \gcd(a, b)^2$$

This is easy to understand intuitively and using the Fundamental Theorem of Arithmetic would be easy but I want to prove it by using the Divisibility axioms and the GCD. I have attempted it and am mentioning my results.

Here $d = \gcd(a, b)$.

$$d | ax + by \implies d^2 | a^2x^2 + 2axby + b^2y^2 \implies d^2 | a^2x^2 + b^2y^2 \implies \gcd(a^2, b^2) | d^2$$

I feel that this is not necessary. Please help.

Also, the proof can be generalized to $$\gcd(a^n, b^n) = \gcd(a, b)^n$$

Any help would be appreciated.

Answer 1

If $d=\gcd(a,b)$, then $a=da_1$ and $b=db_1$ for some $a_1,b_1$ relatively prime.

Then $a^2=d^2a_1^2$ and $b^2=d^2b_1^2$.

Assume there is a prime $p$ which divides $a_1^2$ and $b_1^2$. Since $p$ is prime, $p\mid a_1$ and $p\mid b_1$ (if a prime divides a power, then it divides the number - basic property of primes). Contradiction, because $a_1,b_1$ relatively prime.

Hence $d^2$ is the GCD of $a^2$ and $b^2$.

To have a general proof, just replace the exponent "2" with "n".

Answer 2

GCD is

$\prod_{i=1}^np_i^{min(e_i)}$ where $p_i$ is common prime factor of the two numbers and $min(e_i)$ is the smaller exponent of the $2$.

$\prod_{i=1}^np_i^{2min(e_i)}=\prod_{i=1}^np_i^{min(e_i)}p_i^{min(e_i)}=\prod_{i=1}^np_i^{min(e_i)}\prod_{i=1}^np_i^{min(e_i)}=(\prod_{i=1}^np_i^{min(e_i)})^2$