Question

find all the solution to

p^x = y^4 +4

where p is a prime
and x and y r natural numbers,

Answer 1

seriously

Answer 2

\[p^x = (y^2 - 2y + 2) (y^2 + 2y +2)\]

Answer 3

ok

Answer 4

\[y^2 -2y +2 = p^a\]

\[y^2 + 2y + 2 = p^b\]

a + b = x

Answer 5

now

Answer 6

Lets think together. Taking from where moneybird stopped, we can find that:
\(y=1\pm \sqrt{p^a-1}\) or \(y=-1\pm\sqrt{p^b-1}\).

Answer 7

For solutions over the natural numbers \(p^a-1\) and \(p^b-1\) have to be complete squares.

Answer 8

An obvious solution occurs at p=5, a=1 and b=0, which gives x=1 and y=1.

Answer 9

so is it p^a - 1 and p^b - 1 must be perfect squares?

Answer 10

Yes, so y will be integer.

Answer 11

Perfect squares yeah! I said complete squares and it's a different thing, sorry!

Answer 12

p=2, x=2 and y=0 is a solution too.

Answer 13

What about this:
\[y^4=(p^{\frac{x}{2}}-2)(p^{\frac{x}{2}}+2)\]

Answer 14

0 is not a natural number so, y = 0 has to be excluded from the solution

Answer 15

the correct answer is only
y=5
x=1
y=1

Answer 16

y has to be odd only

the rhs of the equation turns out to be a multiple of 5 always ,

and hence only one solution exist

Answer 17

Hmm, I did find the only solution after all! YAY!! But I had to prove that no other solution exists! Thanks @Stom anyways!!

Answer 18

u r welcome,

this is an olympiad question,
so shouldve been solved in little time,

but i took a lot of it,

apart from that it was a good experience to solve the question,
and see u guys solving it,

Answer 19

\[p^x=(y^2-2y+2)(y^2+2y+2)\]\[p^a=y^2-2y+2\]\[p^b=y^2+2y+2\]\(b\ge a\) so \(p^a|p^b\) and so does \(y^2-2y+2\ |\ y^2+2y+2\) it is immediate that\[m=\frac{y^2+2y+2}{y^2-2y+2}=1+\frac{4y}{y^2-2y+2}\]must be an integer so\[4y\ge y^2-2y+2\]and \(y\le 5\) ... from here \(y=1,2\) makes \(m\) an integer.

\(y=1\) so \(p^x=5\) and one solution from this \((p,x,y)=(5,1,1)\)

\(y=2\) so \(p^x=20\) impossible