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Question Number 110593 by Aina Samuel Temidayo last updated on 29/Aug/20

$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{possible}\mathrm{integer}\mathrm{n}$$$$\mathrm{such}\mathrm{that}\frac{(\mathrm{n}-1)({\mathrm{n}}^2+\mathrm{n}-3)}{{\mathrm{n}}^2+4}\mathrm{is}\mathrm{also}\mathrm{an}$$$$\mathrm{integer}$$

Answered by floor(10²Eta[1]) last updated on 19/Oct/20

$${\mathrm{n}}^2+4\mid (\mathrm{n}-1)({\mathrm{n}}^2+\mathrm{n}-3)={\mathrm{n}}^3-4\mathrm{n}+3$$$$\iff {\mathrm{n}}^2+4\mid \mathrm{n}({\mathrm{n}}^2+4)-({\mathrm{n}}^3-4\mathrm{n}+3)=8\mathrm{n}-3$$$$\therefore 8\mathrm{n}\equiv 3(\mathrm{mod}{\mathrm{n}}^2+4)$$$$\Rightarrow {64\mathrm{n}}^2\equiv 9(\mathrm{mod}{\mathrm{n}}^2+4)$$$${\mathrm{n}}^2\equiv -4(\mathrm{mod}{\mathrm{n}}^2+4)$$$$\Rightarrow {64\mathrm{n}}^2\equiv -256(\mathrm{mod}{\mathrm{n}}^2+4)$$$$\therefore 9\equiv -256(\mathrm{mod}{\mathrm{n}}^2+4)$$$$\Rightarrow {\mathrm{n}}^2+4\mid 265$$$${\mathrm{n}}^2+4\in \{5,53\}$$$${\mathrm{n}}^2+4=5\Rightarrow \mathrm{n}=\pm 1$$$${\mathrm{n}}^2+4=53\Rightarrow \mathrm{n}=7$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{I}\mathrm{see}.{\mathrm{You}}^{\prime }\mathrm{ve}\mathrm{corrected}\mathrm{your}\mathrm{mistake}.$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}.$$

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