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.