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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≠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]

$$\mathrm{Let}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{be}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{xy}\neq\mathrm{1},\:\mathrm{x}^{\mathrm{2}} \neq\mathrm{y}\:\mathrm{and}\:\mathrm{y}^{\mathrm{2}} \neq\mathrm{x}. \\ $$$$ \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{p}\mid\mathrm{xy}−\mathrm{1}\:\mathrm{and}\:\mathrm{p}\mid\mathrm{x}^{\mathrm{2}} −\mathrm{y} \\ $$$$\mathrm{then}\:\mathrm{p}\mid\mathrm{y}^{\mathrm{2}} −\mathrm{x}\:\mathrm{where}\:\mathrm{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Let}\:\mathrm{p}\:\mathrm{be}\:\mathrm{a}\:\mathrm{prime}.\:\mathrm{Suppose}\:\mathrm{that} \\ $$$$\mathrm{p}\mid\mathrm{x}^{\mathrm{2}} −\mathrm{y}\:\mathrm{and}\:\mathrm{p}\mid\mathrm{y}^{\mathrm{2}} −\mathrm{x},\:\mathrm{must}\:\mathrm{p}\mid\mathrm{xy}−\mathrm{1}? \\ $$$$ \\ $$$$\left[\mathrm{If}\:\mathrm{yes},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{it}.\:\mathrm{If}\:\mathrm{no},\:\mathrm{then}\:\mathrm{give}\:\mathrm{a}\right. \\ $$$$\left.\mathrm{counter}\:\mathrm{example}\right] \\ $$

Answered by Aziztisffola last updated on 29/Aug/20

(i) p∣xy−1 ⇒p∣x(xy−1) ⇒p∣x^2 y−x p∣x^2 −y⇒p∣yx^2 −y^2 ⇒p∣x^2 y−x−yx^2 +y^2 ⇒p∣y^2 −x

$$\:\left(\mathrm{i}\right)\:\mathrm{p}\mid\mathrm{xy}−\mathrm{1}\:\Rightarrow\mathrm{p}\mid\mathrm{x}\left(\mathrm{xy}−\mathrm{1}\right)\:\Rightarrow\mathrm{p}\mid\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{x} \\ $$$$\:\mathrm{p}\mid\mathrm{x}^{\mathrm{2}} −\mathrm{y}\Rightarrow\mathrm{p}\mid\mathrm{yx}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{p}\mid\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{x}−\mathrm{yx}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \Rightarrow\mathrm{p}\mid\mathrm{y}^{\mathrm{2}} −\mathrm{x} \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks but why does p∣x(xy−1) ?

$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{why}\:\mathrm{does}\:\mathrm{p}\mid\mathrm{x}\left(\mathrm{xy}−\mathrm{1}\right)\:? \\ $$

Commented by Aziztisffola last updated on 29/Aug/20

p∣(xy−1)⇒xy−1≡0[p] ⇒x(xy−1)≡0[p] ⇒p∣x(xy−1)

$$\mathrm{p}\mid\left(\mathrm{xy}−\mathrm{1}\right)\Rightarrow\mathrm{xy}−\mathrm{1}\equiv\mathrm{0}\left[\mathrm{p}\right]\: \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{xy}−\mathrm{1}\right)\equiv\mathrm{0}\left[\mathrm{p}\right]\:\Rightarrow\mathrm{p}\mid\mathrm{x}\left(\mathrm{xy}−\mathrm{1}\right) \\ $$

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

$$\left(\mathrm{ii}\right)\:\mathrm{No} \\ $$$$\:\mathrm{p}=\mathrm{3}\:\mathrm{and}\:\mathrm{x}=\mathrm{y}=\mathrm{3} \\ $$$$\:\mathrm{3}\mid\mathrm{3}^{\mathrm{2}} −\mathrm{3}\:\mathrm{and}\:\mathrm{3}\mid\mathrm{3}^{\mathrm{2}} −\mathrm{3}\:\nRightarrow\mathrm{3}\nmid\mathrm{8} \\ $$$$\mathrm{or}\:\mathrm{p}=\mathrm{2};\:\mathrm{x}=\mathrm{8}\:\mathrm{and}\:\mathrm{y}=\mathrm{4} \\ $$

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