
Integer Solutions
Help!
Show that =  17 has no integer solutions.
(Hint: Show using quadratic reciprocity that y is a square mod 17 by considering each prime
divisor of y. Use this to show 2 is a fourth power mod 17, a contradication.)
Using the fundamental theorem of arithmetic, i assume
By Euler's Criterion (y/17) (mod 17)
I get if any of the exponents are even, then
If any of the exponents are odd, then
This is where i get stuck. (Sorry about the bad notation, i haven't quite figured out LaTeX)

Let be a prime dividing . Then
That is: so 17 is a quadratic residue module p. So it follows that
By the Quadratic Reciprocity Theorem: hence it follows that , that is, all primes dividing are quadratic residues module 17, hence so is .
But clearly, reading the equation module 17 we have: since for some integer , coprime to , it follows that: which is impossible since is not the 4th power of an integer module 17. Thus we conclude that the equation can have no solutions.
 an easy way to check this it to remember that ( being coprime to p) has solutions if and only if , in our case n=4 and p = 17 and we see that 
The prime p cannot be 17, because cannot be divisible by 17, otherwise the equation would imply that 17 also divides . Let and then and note that the LHS is dividible by 17² while the RHS is not.