Let x and y be integers such that xy≠1, x^2 ≠y and y^2 ≠x. (i) Show that p∣xy−1 and p∣x^2 −y then p∣y^2 −x where p is a prime. (ii) Let p be a prime. Suppose that p∣x^2 −y and p∣y^2 −x, must p∣xy−1? [If yes, then prove it. If no, then give a counter example]


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Question Number 110562 by Aina Samuel Temidayo last updated on 29/Aug/20

$Let x and y be integers such that$ $xy \neq 1, x^{2} \neq y and y^{2} \neq x .$ $(i) Show that p ∣ xy - 1 and p ∣ x^{2} - y$ $then p ∣ y^{2} - x where p is a prime .$ $(ii) Let p be a prime . Suppose that$ $p ∣ x^{2} - y and p ∣ y^{2} - x, must p ∣ xy - 1 ?$ $[If yes, then prove it . If no, then give a$ $counter example]$
	Answered by Aziztisffola last updated on 29/Aug/20

	$(i) p ∣ xy - 1 \Rightarrow p ∣ x (xy - 1) \Rightarrow p ∣ x^{2} y - x$ $p ∣ x^{2} - y \Rightarrow p ∣ {yx}^{2} - y^{2}$ $\Rightarrow p ∣ x^{2} y - x - {yx}^{2} + y^{2} \Rightarrow p ∣ y^{2} - x$
	Commented by Aina Samuel Temidayo last updated on 29/Aug/20

	$Thanks but why does p ∣ x (xy - 1) ?$
	Commented by Aziztisffola last updated on 29/Aug/20

	$p ∣ (xy - 1) \Rightarrow xy - 1 \equiv 0 [p]$ $\Rightarrow x (xy - 1) \equiv 0 [p] \Rightarrow p ∣ x (xy - 1)$
	Answered by Aziztisffola last updated on 29/Aug/20

	$(ii) No$ $p = 3 and x = y = 3$ $3 ∣ 3^{2} - 3 and 3 ∣ 3^{2} - 3 ⇏ 3 ∤ 8$ $or p = 2; x = 8 and y = 4$
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