Help!

Show that $\displaystyle 2*y^2$ = $\displaystyle x^4$- 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 $\displaystyle y = {p_1}^{a_1} *....* {p_t}^{a_t}$

By Euler's Criterion (y/17) $\displaystyle \equiv$$\displaystyle y^8$ (mod 17)

I get if any of the exponents are even, then $\displaystyle (\frac{{p_i}^{a_i}}{17}) = 1$

If any of the exponents are odd, then $\displaystyle (\frac{{p_i}^{a_i}}{17}) = (\frac{p_i}{17})$

This is where i get stuck. (Sorry about the bad notation, i haven't quite figured out LaTeX)