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roza2010 wrote: $x^2+xy+y^2\equiv i^2+ij+j^2 (mod 100)$, where $x\equiv i (mod 10), y\equiv j (mod10)$. Unfortunately quite false, for example $x=y=11$, $i=j=1$, and $363 \not \equiv 3\pmod{100}$.

its easy to find that if $2 \mid x^2+xy+y^2 \Rightarrow 2\mid x , 2\mid y$ so $4\mid x^2+xy+y^2$ then for $5$ if $5$ divides one of them it has to divide the other so in this case the problem is solved. if $ x\equiv 1 ,4\pmod{5} $ then $x^2\equiv 1 \pmod{5}$ so $5 \mid 1+y+y^2$ or $5 \mid 1 -y+y^2$ the two are the same and they cant be true! if $x \equiv 2,3 \pmod{5}$ then $x^2 \equiv -1 \pmod{5}$ so $ 5\mid -1+2y+y^2$ or $5 \mid -1-2y+y^2$ the two are the same and they cant be true so $5 \mid y , x$ DONE!

The polynomial $a^2+a+1$ is equivalent to $0$ modulo prime $p$ iff $p \equiv 1 \pmod 3$. Therefore, if $5|x^2+xy+y^2, 5|x,y$ also.

It is given that $x^2+xy+y^2=10m$. Viewing this as a quadratic equation in $y$, the discriminant must be a square: $40m-3x^2=t^2$. Then $2x^2=t^2 \mod 5 \to x=t=0 \mod 5$. Let $x=5x_1,t=5t_1$ and reduce to $8m-15x_1^2=5t_1^2$. So $m=5m_1$ and reduce to $8m_1-3x_1^2=t_1^2$. If $x_1=t_1=1 \mod 2$ then $x_1^2=t_1^2=1 \mod 8$, impossible. So at least one of $x_1, t_1$ are even, which means both are even. Let $x_1=2x_2,t_1=2t_2$ and reduce to $2m_1-3x_2^2=t_2^2$. Taking this $\mod 4$ gives that $m_1$ is even. Thus $m$ is a multiple of $10$ and we are done.

Claim: If $2 \mid x^2 + xy + y^2$, then $4\mid x^2 + xy + y^2$. Proof: Consider some integers $x,y$ with $x^2 + xy + y^2$. Then note that $x^2 + xy + y^2$ being even implies that $x,y$ are both even (as if one of them is odd, say $y$, then the parity is the same as $x^2 + x + 1$, which is odd), so $4$ obviously divides $x^2 + xy + y^2$. $\square$ Claim: If $5\mid x^2 + xy + y^2$, then $25 \mid x^2 + xy + y^2$. Proof: Choose $x,y$ with $5 \mid x^2 + xy + y^2$. Suppose that $25 \nmid x^2 + xy + y^2$, so $x,y$ are both not multiples of $5$ (as one being a multiple of $5$ means the other also is). Note that $5\mid (x-y)(x^2 + xy + y^2) = x^3 - y^3$. Hence $x^3 \equiv y^3 \pmod 5$, so $\left( \frac xy \right)^3 \equiv 1 \pmod 5$. However, since $5 \equiv 2 \pmod 3$, this means $\frac xy\equiv 1 \pmod 5$, so $5\mid x - y$. We then get $5\mid 3x^2$, so $5\mid x$, absurd. $\square$ Therefore if the last digit of $x^2 + xy + y^2$ is $0$, then $2, 5$ divide $x^2 + xy + y^2$, so $4$ and $25$ divide $x^2 + xy + y^2$ also, meaning $100 \mid x^2 + xy + y^2$, so its last two digits are zero.