Math Problem of the Week: When is n^2 + 12n a perfect square

[wiskundehulpnl]

Hello everyone,

Here is this week's problem.

Find all positive integers [tex]n[/tex] such that [tex]n^2 + 12n[/tex] is a perfect square.

Hint: Try to write the expression as a difference of two squares, then think about the factors of a number.

Please post your full solution and explain each step. Different methods are welcome. Good luck!

[Guest]

Hi its Bob,

Nice problem! Let [tex]n^2 + 12n = m^2[/tex]. Completing the square gives [tex](n+6)^2 - m^2 = 36[/tex], so

[tex](n+6-m)(n+6+m) = 36[/tex]

The two factors add up to [tex]2(n+6)[/tex], which is even, so they are both even. The only such factor pairs of [tex]36[/tex] are [tex]2 \times 18[/tex] and [tex]6 \times 6[/tex].

From [tex]2 \times 18[/tex]: [tex]n+6 = 10[/tex], so [tex]n = 4[/tex] and [tex]m = 8[/tex].
From [tex]6 \times 6[/tex]: [tex]n = 0[/tex], which is not positive.

So the only answer is [tex]n = 4[/tex]. Check: [tex]4^2 + 12 \cdot 4 = 64 = 8^2[/tex].