MHB Can You Solve These Number Theory Challenges?

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The discussion centers on solving two number theory challenges. The first challenge involves proving that for positive integers x and y satisfying the equation x^2 = y^2 - 9y, the possible values for x are 6 or 20. This is demonstrated by manipulating the equation and factoring it, leading to a set of conditions that yield these solutions. The second challenge asks to show that for distinct primes p and q, the equation p^(q-1) + q^(p-1) ≡ 1 (mod pq) can be proven using Fermat's Little Theorem. Overall, the thread emphasizes methods and proofs in number theory.
Poirot1
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1)Prove that x,y are positive integers such that $x^2=y^2-9y$, then x=6 or 20.

2) Let p and q be distinct primes. Show that $p^{q-1}+q^{p-1}=1$ (modpq)

Hint for 2) Use Fermats little theorem.
 
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Re: 2 Number theory challenge

Poirot said:
1)Prove that x,y are positive integers such that $x^2=y^2-9y$, then x=6 or 20.
Multiply the equation by $4$ to get $4y^2-4x^2-36y=0$, which can be factorised as $(2y+2x-9)(2y-2x-9) = 81$. The only possibilities are $$2y+2x-9 = \left\{\begin{matrix}1\\ 3\\9 \\ 27 \\ 81 \end{matrix}\right.,\qquad 2y-2x-9 = \left\{\begin{matrix}81\\ 27\\ 9\\ 3 \\ 1 \end{matrix}\right.. $$ Subtract the second of these from the first to get $4x = \left\{\begin{matrix}-80\\ -24\\ \phantom{-1}0\\ \phantom{-}24 \\ \phantom{-}80 \end{matrix}\right..$ Reject the first three cases because $x$ is positive, and we are left with $x = 6$ or $20.$
 
Better method for 1) relies on the following fact: if a and b are coprime positive integers such that ab is a square, then a and b are both squares.
 
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