23^p+29^q=47^r+5^s ?

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

Who can solve the diophantine equation : $23^p+29^q=47^r+5^s$ ?

**Posted:** Tue Jan 18, 2005 1:45 pm

Megus

as 5, 23, 29, 47 are all primes you just have to play with some congruencies to get some result

**Posted:** Wed Jan 19, 2005 1:37 pm

_________________
Przemyslaw Chojecki

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

Can you go further into detail about your congruences? Mr. Green

**Posted:** Wed Jan 19, 2005 2:25 pm

pavel25 · Poincare Conjecture Offline Joined: 15 Jan 2005 Posts: 122

$\displaystyle23+29=47+5$ so one of the answers:
$p=q=r=s=1$

**Posted:** Thu Jan 20, 2005 5:14 am

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

Yes and $(p,q,r,s)=(0,0,0,0)$ too.
But can we solve this equation?

**Posted:** Thu Jan 20, 2005 9:15 am