can the number $x^2 +y^2$ when with x and y positive integers

Question

can the number $x^2 +y^2$ when with $x$ and $y$ positive integers, end in $03$?

I know that $x^2+y^2$ can never end with unit digit 03 but am not sure how would I show the proof of that.

Answer 1

No, it can't. This is because for the sum of two squares to end with the digits $03$ means that the sum is congruent to $3 \mod 100$. Since $4$ divides $100$, the sum must also be congruent to $3 \mod 4$. However, all squares are congruent to only either $0$ or $1$ mod $4$, so the sum of two of them would be congruent, modulo $4$ (i.e., when divided by $4$) to $0$, $1$ or $2$, but never $3$.

Answer 2

Let $x = 10a \pm b$ and $y = 10c \pm d$ where $b,d = 0,1,2,3,4$ or $5$.

So $x^2 + y^2 = 100(a^2 + c^2) + 20(\pm ab \pm cd) + (b^2 + d^2)$.

The only way for $b^2 +d^2$ to end in $3$ is if $b=2$ and $d=3$ (or vice versa)

So $x^2 + y^2 =100(a^2 + c^2) + 20(\pm 2a \pm 3c) + 13$

And the only we for that to end in $03$ is for $20(\pm 2a \pm 3c)+10$ to end in $00$ or for $2(\pm 2a \pm 3c)+1$ to end in $0$.

But $2(\pm 2a\pm 3c) +1$ is odd so that can never happen.