Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer.

Question

$n^2+8n-5$ has to be a perfect square.

How to find all $n$?

Answer 1

$y^2=n^2+8n-5 = (n+4)^2 - 21$ implies $21 = x^2-y^2$, for $x=n+4$.

Write $21 = x^2-y^2= (x-y)(x+y)$. Since $x=n+4 \ge 4$, we must have $x+y=7$ or $x+y=21$.

• $x+y=7$ implies $x-y=3$ and so $x=5$ and $n=1$

• $x+y=21$ implies $x-y=1$ and so $x=11$ and $n=7$

Answer 2

Note that $n^2 + 8x - 5 = (n + 4)^2 - 21$. Now this number will certainly be no square if it is bigger than $(n + 3)^2$ (since there are no squares strictly between $(n + 3)^2$ and $(n + 4)^2$), i.e. if

$$\begin{align*} &n^2 + 8n - 5 > (n + 3)^2 \\ \iff& n^2 + 8n - 5 > n^2 + 6n + 9 \\ \iff& 2n > 14 \\ \iff& n > 7 \end{align*}$$

So we know that $n^2 + 8n - 5$ is no square if $n$ is greater than $7$. But this leaves only $n = 1, 2, \ldots, 7$ to check manually. This shows that $n = 1$ and $n = 7$ are the only numbers for which $n^2 + 8n - 5$ is a square.

Answer 3

Suppose that $$ n^2+8n-5=m^2\tag{1} $$ Then $$ 8n-5=(m-n)(m+n)\tag{2} $$ If $m-n\ge4$, then we have $8n-5\ge4(2n+4)=8n+16$. Thus, $k=m-n\lt4$. It is also obvious that $k=m-n$ must be odd (since $8n-5$ is odd). $$ 8n-5=k(2n+k)=2kn+k^2\tag{3} $$ Thus $$ n=\frac{k^2+5}{8-2k}\tag{4} $$ Plugging $k=1$ and $k=3$ into $(4)$ gives $$ n=1\quad\text{or}\quad n=7\tag{5} $$