Thread: solutions of equations, using rings

My problem is to show that has no solution in the integers.
I've done other problems like this with one variable, where you show that there are no roots in, say or . However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

Thanks a lot!

Originally Posted by kimberu

My problem is to show that has no solution in the integers.
I've done other problems like this with one variable, where you show that there are no roots in, say or . However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

Thanks a lot!

This is probably not the intended method, but it works fine...

Assume towards a contradiction that there are integers with . Then

for some integer .

So , that is, , which implies

for some integer .

So , that is, . But this implies , which is a contradiction.

Originally Posted by kimberu

My problem is to show that has no solution in the integers.
I've done other problems like this with one variable, where you show that there are no roots in, say or . However with this equation I know Z2, Z3, Z4, Z6 won't work and I've found roots in Z5 and Z7 already. Do I need to go even higher or am I missing something obvious? Does anyone know which Zx I should try, if I do indeed need to go higher?

Thanks a lot!

Mod 14 works. However, so does mod 7.

Working mod 7, we want to find m such that .








And so none are congruent to 5 mod 7, and so your equation has no roots...