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Let $a, b, c \in \mathbb{N}$. Show that if $a^2+b^2=c^2$ then $gcd(a,b)=gcd(a,c)=gcd(b,c)$.

Using basic divisibility criteria I couldn't advance anything.

Lucas
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    If the gcd of any two of those variables was $n$, then by algebra you could get the third variable (squared) on one side of the equation, and both sides would be divisible by $n^2$. – Joe Aug 15 '19 at 18:34
  • Note that your criteria isn't always correct: in fact $12^2+16^2=20^2$ is true, but $gcd(12,16)=gcd(12,20)=gcd(16,20)=4$. – Matteo Aug 15 '19 at 20:08
  • @Matteo What do you think is incorrect? – Bill Dubuque Aug 15 '19 at 21:32
  • His thesis isn't always correct: it's yrue only in a primitive term. – Matteo Aug 16 '19 at 06:37

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Hint $\,(a,b)^2\overset{\color{#c00}d}=\, \overbrace{(a^2,b^2) = (a^2,b^2,c^2)}^{\large {\rm by}\ \ c^2\ =\ j\,a^2+k\,b^2}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{j\,a^k+k\,b^2} = (b^2,c^2)}_{\large{\rm by}\ \ a^2\ =\ m\, b^2 + n\, c^2} \overset{\color{#c00}d} = (b,c)^2\,$ via $\,\color{#c00}d = $gcd Freshman's Dream

Shubham Johri
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Bill Dubuque
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