I have a homework problem that asks for a proof that $$p^2+q^2=r^2+s^2+t^2$$ doesn't have a solution where $p,q,r,s,t$ are prime.
I tried considering this equation modulo 2, but that got me nowhere:
If none of the variables are equal to $2$, then the equation does not have a solution, since $p^2+q^2\equiv 0 \ (\textrm{mod}\ 2)$ and $r^2+s^2+t^2\equiv 1 \ (\textrm{mod}\ 2)$. That's a good start. From this thinking we can figure out that in order for the equation to be able to have a prime solution, 1, 3 or all 5 variables have to be even (so, since they're prime, $2$). We can, of course, easily rule out the case where all 5 variables are equal to $2$, since $8\neq 12$. Even the case where 3 of them are equal to $2$ isn't that bad. This case comes down to two subcases: $$p^2+q^2=12$$ for which it is easy to check no solutions exist, and $$p^2+4=8+t^2 \Longrightarrow (p-t)(p+t)=4$$ for which it is again easy to show no solutions exist.
The real problem is the case where only one of these primes is equal to $2$. Again, two subcases exist: $$p^2+q^2=4+s^2+t^2$$ and $$p^2+4=r^2+s^2+t^2$$ I tried moving terms over and factoring, plugging in different values and checking and nothing seems to work. Is my approach wrong or is there a way to make this proof work? Thank you for the help.
Once you work out that there are an odd number of twos, every case is ruled out by working modulo $8$, where the squares of all odd primes are congruent to $1$. It remains to write down the cases: $$4+4\not\equiv4+4+4\\ 1+1\not\equiv4+4+4\\ 1+4\not\equiv1+4+4\\ 4+4\not\equiv1+1+4\\ 1+1\not\equiv1+1+4\\ 1+4\not\equiv1+1+1$$