Number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is of the form $k^{4}$ with $k$ an integer. Find the number of elements $n \in \{1, ..., 100 \}$ such that $n^{4} - 20n^{2} + 100$ is not of the form $k^{4}$ with $k$ an integer.

Notice that 
$$ n^{4} - 20n^{2} + 100 = (n^{2} - 10)^{2} $$
We can find the number of elements such that $(n^{2}-10)^{2} = k^{4}$.
$$(n^{2}-10)^{2} = k^{4} = (k^{2})^{2} \implies n^{2} - k^{2} = 10$$
$$ (n-k)(n+k) = 10 $$
so the possibilities are
$$ n-k = 1, n+k = 10 \implies n, k \notin \mathbb{Z}$$
$$ n-k = 2, n+k = 5 \implies n, k \notin \mathbb{Z}$$
if we swap the cases, same thing also applies. So there is no $n$ such that 
$ n^{4} - 20n^{2} + 100 = k^{4} $
Thus the answer is $100$, all elements in $\{1, ..., 100 \}$.

Is this sufficient? Is there another way to solve this using techniques which are used in contests. Thanks.

Edit :
Also noting Prathyush's answer.
with $(n^{2}-10)^{2} = k^{4}$
we also need to check
$10 - n^{2} = k^{2}$
which means
$$ 10 = n^{2} + k^{2} $$
only solution is $n=1,k=3$ and $n=3, k=1$. So the answer is $98$ elements.

$$n^{4} - 20n^{2} + 100= (n^2-10)^2$$
Thus the condition reduces to finding all $n$ such that $|n^2-10|$ is of form $k^2$. $n=1,3$ satisfy this. Let $n>3$ so that $|n^2-10|=n^2-10$. Suppose there does exist a $k$ satisfying this, then 
$$n^2-10=k^2$$
$$(n-k)(n+k)=10$$
Now the highest power of $2$ in $10$ is $2^1$. Thus if one factor ($n-k$) is even, the other ($n+k$) is odd. But then $n=\frac{(n+k)+(n-k)}{2}$ will not be an integer, which is a contradiction. Thus there doesn't exist any value of $n>3$ which satisfies the equation.
Conclusion: all values of $n$ except $n=1,3$ are such that $n^{4} - 20n^{2} + 100$ is not of the form $k^4$
