Problem
NT-B1-M12-P010 Set 2. Divisor \(n+3\)
#10
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Find all positive integers \(n\) such that \(n+3\mid n^2+2\).
Modulo \(n+3\): \(n\equiv-3\).
We have \(n^2+2\equiv9+2=11\pmod{n+3}\). Hence \(n+3\mid11\). Since \(n\ge1\), \(n+3\ge4\), so \(n+3=11\), \(n=8\). Check works.
Parametric divisibility.