When is $k^4-24k+16$ a perfect square. When is the equation $k^4-24k+16$ perfect square.
(k is an  integer.)
I got this equation as discriminant while solving an equation. I tried to solve it but couldn't i tried to write it in a form of a square but couldnt solve it.I bashed a bit and found 0 and 3 as a solution. Any help would be appreciated. Thanks in advance.
 A: For $k\ge 12$, it can be shown that $(k^2-1)^2\lt k^4-24k+16<(k^2)^2$.
For $k\le-13$, it can be shown that $(k^2)^2\lt k^4-24k+16 < (k^2+1)^2$.
Thus, $k^4-24k+16$ cannot be a perfect square unless
$k\in \{-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11\}$.
Of the elements of that finite set, $k^4-24k+16$ is a perfect square only for $k=-3, 0, $ and $3$.
A: For large $k$, $a=\sqrt{k^4-24k+16}$ will be very close to, but not equal to $k^2$
(and so won't be an integer). How close?
$$|k^2-a|=\frac{|k^4-a^2|}{k^2+a}=\frac{|24k-16|}{k^2+a}\le\frac{|24k-16|}{k^2}.$$
If $|k|\ge 25$, then
$$|24k-16|\le24|k|+16<25|k|\le k^2$$
and so $|k^2-a|<1$. So we only need check $k$ between $-24$ and $24$.
A: This is a refinement of the answers of Angina Seng, J. W. Tanner, and JCAA, using parity (even/odd) to reduce the amount of checking.
Since $-24k+16$ is even, $k^4-24k+16$ has the same parity as $k$.  Thus if $k\ne 0$ and $k^4-24k+16$ is a square, it is either $(k^2-2)^2$ or smaller or $(k^2+2)^2$ or larger.
Suppose first $k$ is positive.  If $(k^2-2)^2 < k^4-24k+16$, we see that $k^4-24k+16$ cannot be a square.  This inequality simplifies to $-4k^2+4<-24k+16$, which further simplifies to $6k<k^2+3$.  This is true as long as $k\ge 6$, so we only need to check $k=1,2,3,4,5$.  Only $k=3$ works.
A similar argument shows we only need to check $k$ from $-6$ to $-1$ for $k$ negative.  Only $k=-3$ works.
A: I think this is a correct solution. Suppose that $k^4-24k+16$ is a square.

*

*Suppose that $-24k+16\ge 0$. Then $k^4-24k+16\ge (k^2+1)^2=k^4+2k^2+1$, so $-24k+16\ge 2k^2+1$ or $2k^2+24k-15\le 0$ or $k^2+12k-7.5\le 0$. This means $$-6-\sqrt{36+7.5}\le k\le -6+\sqrt{36+7.5}.$$
Which means $-13\le k\le 0$ (remember that $-24k+16>0$). Checking all these $k$ we get that only $k=-3,0$ are good.


*Suppose that $-24k+16<0$ (in particular $k\ge 0$). Then $k^4-24k+16\le (k^2-1)^2=k^4-2k^2+1$, so $2k^2-24k+15\le 0$, or $k^2-12k+7.5\le 0$. Thus $6-\sqrt{36-7.5} \le k\le 6+\sqrt{36-7.5}$. So $1\le k\le 11$. Checking all these $k$, we get $k=3$.

Thus there are three possibilities: $k=-3,0,3$.
A: Let's write the expression as $k^4 - 24k + 16 = (k^2 \pm 4)^2 = k^4 \pm 8k^2+16$.
Given the first and last term are same in both, the second term has to be equal for it to be perfect square which leads to -
$\pm8k^2 = -24k \implies k (k \pm 3) = 0$ which gives $k = 0, -3, 3$.
