Mathematics Asked by Henry Cai on December 2, 2020

how do we prove that if $a^2+b^2=c^2$, and they are all integers, there would be at least one of them to a multiple of 5.

What I did is to assume that none of them to be multiple of 5. So$$a=5p_1+r_1, b=5p_2+r_2, c=5p_3+r_3, 0<r<5$$.

By direct substitution back to the equation, we would have:

$$5(5p_1^2+2p_1r_1+5p_2^2+2p_2r_2-5p_3^2-2P_3r_3)=r_3^2-r_1^2-r_2^2$$.

Then, we can list all possible scennarios to show that for this equality to be held, there must be at least 1 r to be 0. I want to ask is there any quick way we can show $$r_3^2-r_1^2-r_2^2$$is a multipler of 5 unless one of them to be 0. Thank you.

Modulo $5$, all squares are in ${-1,0,1}$. The only ways two squares can add up to a third square modulo $5$ are $$0+0=0$$ $$1+0=1$$ $$-1+0=-1$$ $$1+(-1)=0$$ There is a zero in all four possibilities. So in $a^2+b^2=c^2$, at least one variable is a multiple of $5$.

Correct answer by Parcly Taxel on December 2, 2020

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