Can You Solve This Challenging Number Theory Problem?

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SUMMARY

The discussion centers on the number theory problem involving the equation $(n-1)x^2 - px + n = 0$, which requires two positive integer solutions where both $n$ and $p$ are natural numbers. The main objective is to prove that $p^p - n^n = 23$. Participants explore various approaches to demonstrate the relationship between the parameters and the specified equation, ultimately aiming to establish the validity of the claim through rigorous mathematical reasoning.

PREREQUISITES
  • Understanding of quadratic equations and their properties.
  • Familiarity with number theory concepts, particularly integer solutions.
  • Knowledge of natural numbers and their characteristics.
  • Experience with mathematical proof techniques.
NEXT STEPS
  • Study the properties of quadratic equations in number theory.
  • Research methods for finding integer solutions to polynomial equations.
  • Explore the implications of the equation $p^p - n^n = 23$ in number theory.
  • Learn about mathematical proof strategies, particularly in the context of integer equations.
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Mathematicians, number theorists, and students interested in solving complex equations and understanding integer solutions in polynomial contexts.

Albert1
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given equation: $(n-1)x^2-px+n=0 $ has two positive integer solutions , (here $n,p \in N$)

prove :$p^p-n^n=23$
 
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Albert said:
given equation: $(n-1)x^2-px+n=0 $ has two positive integer solutions , (here $n,p \in N$)

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

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 $p^p-n^n = 3^3-2^2 = 23$
 

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