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)
Jul 31st 2009, 10:18 AM
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.