Suppose $\sqrt2=a/b$, with $gcd(a,b)=1$. Then $3|(a^2+b^2)$ implies that $3|a$ and $3|b$,

Question

Suppose $\sqrt2=a/b$, with $\gcd(a,b)=1$. Then $a^2=2b^2$, so that $a^2+b^2=3b^2$. But $3|(a^2+b^2)$ implies that $3|a$ and $3|b$, a contradiction.

I don't understand how $3|(a^2+b^2)$ implies that $3|a$ and $3|b$. I'd appreciate any explanation to this.

Answer 1

(Nearly complete) hint: The only perfect squares $\mod 3$ are $0$ and $1$.

Answer 2

We can write $a=3p+q$ and $b=3p'+q'$, with $q,q'=0,1$ or $2$. Then $$ a^2+b^2=9(p^2+p'^2)+6(pq+p'q')+q^2+q'^2 $$ So if $3|a^2+b^2$, we must have $3|q^2+q'^2$. Then you can just check case by case that the only way this can happen is when $q=q'=0$ (there are only three choices each for $q,q'$, and only 6 distinct cases up to symmetry).