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?

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- Apr 3rd 2011, 05:42 PMAdamCDiophantine 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 PMUnbeatable0
No need for induction. Try to guess a solution which will be valid for all $\displaystyle n$ (and dependent on $\displaystyle n$), using a single solution to $\displaystyle x^2+y^2=17$.

(Sorry if this hint was too big) - Apr 5th 2011, 01:49 AMSalome
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 AMtonio

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

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

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

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

Tonio - Apr 5th 2011, 09:11 AMOpalg
- Apr 5th 2011, 02:31 PMtonio