Find all triples $(x, y, p)$ $(x,y,p)$ where $x$ $x$ and $y$ $y$ are positive integers and $p$ $p$ is a prime, satisfying the equation $x^{5} + x^{4} + 1 = p^{y}$ $x^5+x^4+1 = p^y$ ?

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Answer 1 · 2017-08-15T18:20:51

See below.

$x^{5} + x^{4} + 1 = (x^{3} - x + 1) (x^{2} + x + 1)$ $x^5+x^4+1 = (x^3-x+1)(x^2+x+1)$ so

$(x^{3} - x + 1) (x^{2} + x + 1) = p^{y}$ $(x^3-x+1)(x^2+x+1)=p^y$ then

${(x^3-x+1=p^(y-z)),(x^2+x+1=p^z):}$

or

${(x^3-x+1=p^(y-z)),(x^3-1=(x-1)p^z):}$

with $0 le z le y$

subtracting the first from the second we have

$x-2=(x-1)p^z-p^(y-z)$ or

$x = 1/(p^z-1)(p^z+p^(y-z)-2)$

or

$x=1+(p^(y-z)-1)/(p^z-1)$ here we have integer solutions for

$y-z=k z$ or

$z =y/ (k+1)$

for $k = 0,1,2,3,cdots$

EXAMPLES

Supposing that

$y=7$ we have solutions for $k = {0, 6} rArr z = {7,1}$
$y = 8$ we have solutions for $k = {0,1,3,7} rArr z = {8,4,2,1}$
etc.

Find all triples (x,y,p)(x,y,p)(x,y,p) where xxx and yyy are positive integers and ppp is a prime, satisfying the equation x^5+x^4+1 = p^yx5+x4+1=pyx^5+x^4+1 = p^y?