# Diophantine eq

• Apr 3rd 2011, 05:42 PM
Diophantine eq
Thanks heaps but there is another one which I cant figure out. I have to prove that for each positive integer n there exist integers x,y,z > 1 so that:
x^2+y^2=17z^(4n)

:S

How does induction work with 4 variables?
• Apr 3rd 2011, 10:42 PM
Unbeatable0
No need for induction. Try to guess a solution which will be valid for all $n$ (and dependent on $n$), using a single solution to $x^2+y^2=17$.

(Sorry if this hint was too big)
• Apr 5th 2011, 01:49 AM
Salome
I think AdamC and I are in the same class ;-)

The question asks us specifically to prove the proposition using simple induction. But, Unbeatable0, if I've taken your hint correctly, the values for x, y & z need to be greater than 1.
• Apr 5th 2011, 03:55 AM
tonio
Quote:

Originally Posted by AdamC
Thanks heaps but there is another one which I cant figure out. I have to prove that for each positive integer n there exist integers x,y,z > 1 so that:
x^2+y^2=17z^(4n)

:S

How does induction work with 4 variables?

Why do you worry about the variables? Worry about n...

For $n = 1:\,\,4^2+1^2=17\cdot 1^{4}$...check

Suppose for $n:\,\,\exists x,y,z\,\,s.t.\,\,x^2+y^2=17z^{4n}$ , and we shall prove for n+1. But

$z^{4(n+1)}=z^4z^{4n}$ , so using the inductive hypothesis...(Giggle)(Wink)

Tonio
• Apr 5th 2011, 09:11 AM
Opalg
Quote:

Originally Posted by tonio
Why do you worry about the variables? Worry about n...

For $n = 1:\,\,4^2+1^2=17\cdot 1^{4}$...check

Here, you should worry about the variables, because the question specifies integers x,y,z > 1. (Evilgrin) (But it's not hard to adapt the base case example to satisfy those conditions.)
• Apr 5th 2011, 02:31 PM
tonio
Quote:

Originally Posted by Opalg
Here, you should worry about the variables, because the question specifies integers x,y,z > 1. (Evilgrin) (But it's not hard to adapt the base case example to satisfy those conditions.)

Good point. Completely missed that bigger than 1 thingy.

Tonio