Key idea. Complete squares so that the difference has a fixed value.

1. Write the required product as \(m(m+1)\), where \(m\) is a positive integer. Our equation is \(n^2+2n+12=m(m+1)\).

2. Multiply by 4 and complete squares:

   \[(2m+1)^2-[2(n+1)]^2=45.\]

3. Let \(u=2m+1-2(n+1)\), \(v=2m+1+2(n+1)\). Then \(uv=45\). The second factor is positive, so the first is positive too; also \(u<v\). Thus the possible pairs are \((1,45),(3,15),(5,9)\).

4. Add and subtract the definitions to recover

   \[m=\frac{u+v-2}{4},\qquad n=\frac{v-u}{4}-1.\]

   This gives

   \(u\)	\(v\)	\(m\)	\(n\)
   \(1\)	\(45\)	\(11\)	\(10\)
   \(3\)	\(15\)	\(4\)	\(2\)
   \(5\)	\(9\)	\(3\)	\(0\)

5. Exclude \(n=0\), because n must be positive. For \(n=2\), the expression is \(20=4\cdot5\); for \(n=10\), it is \(132=11\cdot12\). Both remaining cases work.

Remember. A finite list of divisor pairs can replace a search over infinitely many integers.