For how many natural numbers $a$ is the expression $\sqrt{\frac{a+64}{a-64}}$ also a natural number? I cannot seem to do this without using trial and error.
$\endgroup$ 12 Answers
$\begingroup$If $\sqrt{\frac{a+64}{a-64}}=n$, then $a+64=n^2(a-64)$, i.e. $(n^2-1)a=64(n^2+1)$, thus $n^2-1\mid 64(n^2+1)$. We need to distinguish two cases:
$n$ is even. In that case, $n^2-1$ and $n^2+1$ are both odd, and are, moreover, coprime (*). Thus, $n^2-1$ is an odd number dividing $64$, i.e. $n^2-1=\pm 1$. The only solution there is $n=0, a=-64$ - but this is not a natural number.
$n$ is odd. In that case, $n^2+1\equiv 2\pmod 4$ (**), so $n^2-1\mid 128\cdot\frac{n^2+1}{2}$ where $\frac{n^2+1}{2}$ is odd, and, similarly to the previous case, $n^2-1$ is coprime with $\frac{n^2+1}{2}$(*), so $n^2-1$ is an even number dividing $128$. Thus, the candidates for $n^2$ are $3, 5, 9, 17, 33, 65, 129$. Only $n^2=9$ gives an integer $n=3$, which gives the solution $a=80$.
Conclusion: the only such number is $a=80$.
(*) In the first part, any divisor of $n^2-1$ and $n^2+1$ would need to divide their difference, i.e. $2$, but $2$ is not a divisor as those numbers are odd. In the second part, any divisor of $n^2-1$ and $\frac{n^2+1}{2}$ would need to divide $n^2+1$ as well, so again it can only be $2$, but $2$ does not divide $\frac{n^2+1}{2}$.
(**) $(2k+1)^2+1=4k^2+4k+2$
$\endgroup$ $\begingroup$If $\sqrt{\frac{a+64}{a-64}}=n$, then $1 + \frac{128}{a-64} = n^2$ and so $a-64$ divides $128$.
Therefore, $a=64\pm 2^e$, with $e \in \{0,1,\dots,7\}$. Only $16$ candidates to test.
The only solution with $a\ge 0$ and $1 + \frac{128}{a-64}$ a square is $a=80$ and $n=3$.
$\endgroup$ 2
