2021/030)Find prime p and positive integers x,y such that $p^x = y^4 + 4$

We get factoring RHS 

$p^x = y^4 + 4 = (y^4 + 4y^2 + 4) - 4y^2$

$= (y^2 - 2y +2)(y^2+ 2y2)$

as RHS is a power of p so each factor is power of p

and clearly $ (y^2 - 2y +2) < (y^2+ 2y+2)$

so $(y^2 - 2y +2) |  (y^2+ 2y+2)\cdots(1)$

Or $(y^2 - 2y +2) |  (y^2 - 2y +2) -  (y^2+ 2y+2)$

Or $(y^2 - 2y +2) |  (y^2 - 2y+2) -  (y^2+ 2y+2)$

Or $(y^2- 2y+2 | 4y\cdots(2)$

Hence $(y^2- 2y+2 | 4y^2\cdots(3)$

From (1) and (3) we get

$(y^2 - 2y +2) |  4(y^2+ 2y+2)- 4y^2$

or $(y^2 - 2y +2) |  8y+8 $

$(y^2 - 2y +2) |  8(y + 1)\cdots(4) $

As y and y+ 1 are co-primes from 2) ad (4)

$(y^2 - 2y +2) |  8\cdots(4) $

 

So we have 4 cases $y^2-2y+2$ = 1 or 2 or 4 or 8

Let us take the csses one by 1

case 1)

$y^2-2y+2 = 1$

or $y^2-2y+1= 0$ or $y= 1$

this gives $p^x = 1^4 + 4 = 5$ giving $p=5,x = 1$

 Case 2)

 

$y^2-2y+2 = 2$

or $y^2-2y = 0$ or $y= 0$ or 2 

$y=2$ gives $p^x = 20$ which s not a power of a prime

Case 3)

 

$y^2-2y+2 = 4$

or $y^2 - 2y - 2= 0$ this does not have integer solution

Case 4)

 

$y^2-2y + 2 = 8$

or $y^2 - 2y - 6= 0$ and this does not have integer solution

So the solution set $p=5,x=1,y=1$