understood question wrong

$a \vdots b$ means that $a$ is divisible by $b$ or $b$ divides $a$, equivalently $b\mid a$.

We have $$\left(\frac{-25}{p}\right)=1,$$where $p\neq 5$. Note that $$\left(\frac{-25}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{5^2}{p}\right)=\left(\frac{-1}{p}\right)$$This is $-1$ if $p\equiv 3 \pmod 4$. Notice that there are $(x,y)=(5,2)$, where its prime factors are only $p\equiv 1 \pmod 4$. Idk how to proceed from here?

@parmenides51, does there exist a official solution by Belarusian Math Jury, etc? If so, can you share it (or the source)

we may await for others, to try it, if in a few days noone responses I shall send the solution

looking it at it $(mod 503)$, we have $2011^x-1007^y=(-1)^x-1^y (mod503) $ as discussed above $503=3 (mod4)$ so it cant divide $2011^x-1007^y$ And so x must be odd now we look at it $(mod 3)$ $2011^x-1007^y=1^x-(-1)^y$ (mod3) and so y must be odd Finally looking at it $(mod4)$, we have $(-1^x)-(-1)^y=0 (mod4)$ as the parity and x and y are equal however $max(V_2(m^2+25))=1$ and so there is a contradiction No Solutions PS Thank you @rafaello for the 3(mod4) result

@above, seems legit ig. I still would like to see official sol.

here is the official one

Attachments:

I also wanted to solve it with same way.. But I wasn't able to prove that there's always a prime of form $4k+3$ which divides $2011^x-1007^y$ So what's the motivation behind taking $\mod 503$ ?

The motivation of taking mod 503 is because $503| 2011+1$ and $503|1007-1$

Hint:take mod 503 first and ur second move for questions is mod 4