Found this one in one of my old test papers, give it a try folks:

If we have a prime no $ p $ and natural nos $ x $ and $ y $ such that $ p^x = y^4 + 4 $,
then how many such ordered triplets $ (p, x, y) $ exist?

Question

Found this one in one of my old test papers, give it a try folks:

If we have a prime no $ p $ and natural nos $ x $ and $ y $ such that $ p^x = y^4 + 4 $,
then how many such ordered triplets $ (p, x, y) $ exist?

dumbcow · Answer

i have no idea :)

katrinakaif · Answer

Why is everyone looking through their old papers now? :P

anonymous · Answer

across question reminded me I have to work on mine too :P

anonymous · Answer

no is "no" and nos is numbers.

anonymous · Answer

I think it has only one solution which is $(5,1,1)$.

anonymous · Answer

@AnwarA: The thing is .. you have to prove it ;)

anonymous · Answer

I know! Workin' on it ;)

anonymous · Answer

Cool! :)

jhonyy9 · Answer

$$(2n+1)^{x}=y ^{4}+4$$

jhonyy9 · Answer

- what is your opinion(s) from this ?

xln(2n+1)=4ln(y)+ln4
xln(2n+1)=4ln(4y)

- so i think this is equal when x=4 and 2n+1=4y so y=(2n+1)/4 so but we know that 
(2n+1) not can divide by 4 never because every primes grater than 2 are odd

mathmate · Answer

ln(y^4+4) does not equal 4ln(y)+ln(4)

anonymous · Answer

http://openstudy.com/study#/updates/4eeabc44e4b0367162f501ba