we'll prove that for every $ m$ not divisible by $ 107$ there exist $ x,y$ such that $ 107|x^2 + xy + y^2 + m$ and $ x - y$ is not divisible by $ 107$... we will consider everything modulo 107. if $ m$ is a nonzero quadratic residue, we have that $ m\equiv p^2\equiv 100(p\cdot 10^{ - 1})^2\equiv 100r^2$, for some positive integers $ p,r$ less than $ 107$... take $ (x,y) = (2r,r)$ and we're done... if $ m$ is not a quadratic residue, we have that it can be written as $ m\equiv 7p^2$... we have that $ 13$ and $ 100$ are quadratic residues, so the congruence $ 13t^2\equiv 100p^2$ has a (nonzero) solution for all $ m$... take $ (x,y) = (3t,t)$ and we're done...

if $ m$ is not a quadratic residue you can use the fact that $ -1$ is not an RC mod $ 107=4k+3$ Then if $ m\equiv -t^2$ $ (mod 107)$ Then if you put $ x=t$ and $ y=0$ $ (mod 107)$ you arrive to the desired result In the case $ m$ is an RC I have done something similar to you Daniel

I'm sorry but can you please explain your solution, I don't get $ m\equiv p^2\equiv 100(p\cdot 10^{ - 1})^2\equiv 100r^2$

m is a quadratic residue, so let, $ m\equiv p^2$... also 10 and 107 are coprime, so there exists an integer $ t$ such that $ 10t\equiv 1$, which can be denoted as $ 10^{-1}$... so $ m\equiv p^2\equiv p^2(10t)^2\equiv 100(pt)^2$, and that's it... is it clear now?

yes it is thank you very much

campos wrote: if $ m$ is not a quadratic residue, we have that it can be written as $ m\equiv 7p^2$... we have that $ 13$ and $ 100$ are quadratic residues Can someone explain both of these claims?

I have an other solution ,not very good but I think it is useful . My solution also true for all prime p of the form $ 12k - 1$ . Suppose the contradiction , $ m$ is not a multiple of $ p$ . The first we will show that there isn't any pair of $ (x,y)$ such that $ p|x^2 + xy + y^2 + m = f(x,y)$ . If not , we can find two integers $ x,y$ such that $ p$ is a divisor of $ x^2 + xy + y^2 + m$ . Because $ f(x,y) = f( - y - x,y)$ so $ p$ is also divisible $ f(x, - y - x)$ . By the condition of the problem we must have $ y\equiv x\equiv - y - x (mod p)$ ,therefore $ x\equiv y\equiv 0 (\mod p)$ . Other hand $ p|f(x,y)$ so $ p|m$ ,which gives contradiction . Because $ p$ is not a divisor of $ f(x,y)$ for all pairs of integer $ (x,y)$ so $ - 3y^2 - 4m$ is a non quadratic reside mod p for all integers $ y$ . Which means that : $ \binom{ - 3y^2 - 4m}{p} = - 1$ ,but p has form $ 4k + 3$ so $ \binom { - 1}{p} = - 1$ , hence $ \binom{3y^2 + 4m}{p} = - 1$ . It follows that $ \{3y^2 + 4m,y = 1,..,\frac {p - 1}{2}\}$ be the set of all quadratic reside mod p . Other hand , because $ \binom{3}{p} = 1$ so $ \{3y^2,y = 0,..,\frac {p - 1}{2}\}$ is the set of all quadratic reside mod p . Hence some of numbers in each set is congruent to each other to modulo p ,so $ 2(p - 1)m\equiv 0 (\mod p)$ ,or it is equivalent $ m\equiv 0 (\mod p )$ (contradiction ) . This problem has been proved .

dgreenb801 wrote: campos wrote: if $ m$ is not a quadratic residue, we have that it can be written as $ m\equiv 7p^2$... we have that $ 13$ and $ 100$ are quadratic residues Can someone explain both of these claims? Are these claims correct?

i'm sorry i haven't answered before... yes, i think the claims are correct... suppose that $ 7p^2\equiv q^2$ for some nonzero $ p$ and some integer $ q$... then, this would imply that $ 7$ is a quadratic residue (taking the inverse of $ p^2$), which is false... the claims for $ 13$ and $ 100$ are true... in fact, when solving these problems i tried to simulate the competition... i took my 4 1/2 hours, and since i had enough time, i calculated (by hand) all the quadratic residues modulo $ 107$...