+302 If n is a positive integer, what is the sum of all possible values of n for which \(n^2\)+72 is a perfect square?
+9673 Suppose that \(n^2 + 72 = m^2\), where m is a positive integer.
Moving terms around gives \(m^2 - n^2 = 72\), which in turn means \((m - n)(m+n) = 72\).
Note that m - n and m + n have the same sign (since they differ by an even number).
They cannot be both odd, because their product is an even number. Hence they must be both even.
The positive factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
So, we get the possible values \(\begin{cases} m + n = 36\\ m - n = 2 \end{cases}\), \(\begin{cases} m + n = 18\\ m - n = 4 \end{cases}\), \(\begin{cases} m + n = 12\\ m - n = 6 \end{cases}\).
The rest should be simple enough to figure out.
+302 Pluging in random numbers for n works, but can anyone find a better way to get the possible values of n? Thank you
+9673 Suppose that \(n^2 + 72 = m^2\), where m is a positive integer.
Moving terms around gives \(m^2 - n^2 = 72\), which in turn means \((m - n)(m+n) = 72\).
Note that m - n and m + n have the same sign (since they differ by an even number).
They cannot be both odd, because their product is an even number. Hence they must be both even.
The positive factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
So, we get the possible values \(\begin{cases} m + n = 36\\ m - n = 2 \end{cases}\), \(\begin{cases} m + n = 18\\ m - n = 4 \end{cases}\), \(\begin{cases} m + n = 12\\ m - n = 6 \end{cases}\).
The rest should be simple enough to figure out.
