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

Question

1666 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-15T18:20:51

See below.

#x^5+x^4+1 = (x^3-x+1)(x^2+x+1)# so

#(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.