diophantine equation

blang · Poincare Conjecture Offline Joined: 13 Jan 2005 Posts: 157

Let p,q,r different primes.
Prove that the diophantine equation: z^r +x^p=2y^q has an infinite set of solutions with x>1, y>1 and z>1.

**Posted:** Thu Jan 13, 2005 12:56 pm

ondrob

Put:
$z=a^{pq}$
$x=a^{qr}$
$y=a^{rp}$
And for each $a>1$ you have solution => infinitely many solutions.

**Posted:** Thu Jan 13, 2005 1:35 pm

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Mrzne tu, ni příští, i chlebem s nedochodím.
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blang · Poincare Conjecture Offline Joined: 13 Jan 2005 Posts: 157

Ok it's obvious ! Blush

A boob for my first post...Well done!
In fact, I have forgotten the following condition:
z^r+x^p=2y^q with z^r and x^p different.
___________
blang

**Posted:** Fri Jan 14, 2005 9:33 am

pbornsztein

**Posted:** Fri Jan 14, 2005 9:57 am

blang · Poincare Conjecture Offline Joined: 13 Jan 2005 Posts: 157

Nobody's interested by this problem? Smile

(corrected)
Let $p,q,r$ different primes.
Prove that the diophantine equation: $z^r +x^p=2y^q$ has an infinite set of solutions with $x>1$ , $y>1$ , $z>1$ and $z^r\neq x^p$ .

**Posted:** Sun Jan 30, 2005 6:23 am

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blang (sorry for my french english!)

Myth

Let $u$ be such integer that $q\mid pru+1$ , $t$ be an arbitrary odd integer number, $w=\frac{t^r+1}{2}$ .
Set $x=w^{ru}$ , $z=w^{pu}t$ , $y=w^{(pru+1)/q}$ . Then
$x^p+z^r=w^{pru}+w^{pru}t^r=w^{pru}(1+t^r)=w^{pru}\cdot 2w=2y^q.$

Hope, it is correct.

**Posted:** Sun Jan 30, 2005 7:24 am

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Myth is out of here

blang · Poincare Conjecture Offline Joined: 13 Jan 2005 Posts: 157

Well done, Myth.
Other solution:
By chinese lemma, we can find infinitely many $(\alpha,\beta)\in\mathbb{N}^2_*$ such that:
$\alpha\equiv -1[q]$ , $\alpha\equiv 0[pr]$ and
$\beta\equiv -1[r]$ , $\beta\equiv0[pq]$ .
Thus, $(x,y,z)=\left (2^{\frac{\alpha}{p}}3^{\frac{\beta}{p}},2^{\frac{1+\alpha}{q}}3^{\frac{\beta}{q}} ,2^{\frac{\alpha}{r}}3^{\fra...$ is a solution.

**Posted:** Sun Jan 30, 2005 12:24 pm

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blang (sorry for my french english!)