MHB Can You Solve This Challenging Number Theory Problem?

[Albert1]

given equation: $(n-1)x^2-px+n=0 $ has two positive integer solutions , (here $n,p \in N$)

prove :$p^p-n^n=23$

 

[kaliprasad]

given equation: $(n-1)x^2-px+n=0 $ has two positive integer solutions , (here $n,p \in N$)

prove :$p^p-n^n=23$

Spoiler

because both are integers so the coefficient of $x^2$ after dividing by common factor should be 1
so
$(n-1) | n $ => $(n-1) | 1$ so n = 0 or 2 but n cannot be zero so 2
and $(n-1) | p$ which meets criteria
so we have
$x^2-px + 2 = 0$
it has 2 positive roots so they must be 1 and 2 so equation is $x^2-3x + 2 = 0$
so p = 3
hence $p^p-n^n = 3^3-2^2 = 23$