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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.

Martin Sleziak
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TheRandomGuy
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  • you can prove that $ d^2 / gcd(a^2, b^2) $ by observing that $a^2 = s.d^2 $ and $b^2 = t.d^2$ for some natural numbers s and t. Hint: $d = gcd(a,b)$ – inquisitive Mar 03 '16 at 14:10
  • @inquisitive Sorry. But how do I show that? – TheRandomGuy Mar 03 '16 at 14:13
  • @inquisitive So you mean $$d|a \implies d^2 | a^2 \implies sd^2 = a^2$$ – TheRandomGuy Mar 03 '16 at 14:18
  • since $d = gcd(a,b) , a = c.d $ and $ b=e.d $ for some natural numbers $ c $ and $ e$ so, $a^2 = c^2d^2$ and $ b^2 = e^2d^2 $ – inquisitive Mar 03 '16 at 14:18
  • yes you got it! – inquisitive Mar 03 '16 at 14:19
  • You could also write $a=p_1^{e_1}\cdots p_r^{e_r} a'$ and $a=p_1^{s_1}\cdots p_r^{f_r}b'$, where gcd$(a',b')=1$ and $p_1,\ldots,p_r$ are the primes in common, then gcd$(a,b)=p_1^{\min{e_1,f_1}}\cdots p_1^{\min{e_r,f_r}}$. If you take the squares then... – PITTALUGA Mar 03 '16 at 14:22
  • @inquisitive $d^2 | a^2x^2 + 2axby + b^2y^2 \implies d^2 | a^2x^2 + b^2y^2$ I doubt this step. I took it because I knew that both a and b were multiples of d hence $d^2|ab$. But if we are talking about higher n's then how should I do that? – TheRandomGuy Mar 03 '16 at 14:24
  • this is correct since $d^2| a^2x^2 + 2axby + b^2y^2 $ means $\exists $ some $ w $ s.t $d^2.w = a^2x^2 + b^2y^2 + 2axby $ but since $ab = fd^2$ ,$ a^2x^2 + b^2y^2 = (w-f)d^2$ – inquisitive Mar 03 '16 at 14:32

2 Answers2

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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".

tong_nor
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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 $2$.

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

miniparser
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