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Find the product of all integer values of $n$ such that

n^{2} - 24 n + 147

is a square number.

Hint

Assume that $m, n \in Z$ and

n^{2} - 24 n + 147 = m^{2}

Complete the square on the left-hand side.

Hint

Aim to get your equation in the form

a^{2} - b^{2} = c

and factorise the left-hand side into two brackets.

What can you conclude from the possible factorisations of $c$ into two integers?

Solution

Assume that $m, n \in Z$ and

\begin{aligned} n^{2} - 24 n + 147 & = m^{2} \\ (n - 12)^{2} + 3 & = m^{2} \\ m^{2} - (n - 12)^{2} & = 3 \\ (m - n + 12) (m + n - 12) & = 3 \end{aligned}

Notice that $3$ has been factored into two integers. There are four possible cases:

Case 1

\begin{aligned} m - n + 12 & = 1 \\ m + n - 12 & = 3 \end{aligned}

In this case, $m = 2, n = 13$ .

Case 2

\begin{aligned} m - n + 12 & = 3 \\ m + n - 12 & = 1 \end{aligned}

In this case, $m = 2, n = 11$ .

Case 3

\begin{aligned} m - n + 12 & = - 1 \\ m + n - 12 & = - 3 \end{aligned}

In this case, $m = - 2, n = 11$ .

Case 4

\begin{aligned} m - n + 12 & = - 3 \\ m + n - 12 & = - 1 \end{aligned}

In this case, $m = - 2, n = 13$ .