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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.

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$$n^{4} - 20n^{2} + 100= (n^2-10)^2$$ 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 $k^4$

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  • $\begingroup$ Thanks, +1. But one thing, your statement $(n-k)$ even implies $(n+k)$ odd is not accurate. $n-k$ even, then $(n+k) = (n-k)+(2k) = 2m+2k$ is also even. $\endgroup$ – Arief Anbiya Apr 1 '18 at 9:40
  • $\begingroup$ @Arief But according to the equation$(n-k)(n+k)=10$, if one is even the other must be odd. Hence the contradiction, and there cannot exist any values of $n$ satisfying it. $\endgroup$ – Prathyush Poduval Apr 1 '18 at 9:41
  • $\begingroup$ @arief this is exactly the contradiction. You get $n+k$ is even because $n-k $ is even and also $n+k $ odd since $2||10$. $\endgroup$ – Or Kedar Apr 1 '18 at 9:48
  • $\begingroup$ @PrathyushPoduval Understand.. $\endgroup$ – Arief Anbiya Apr 1 '18 at 10:03
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$$n^{4} - 20n^{2} + 100=$$

$$ (n^2-10)^2=k^4 $$

$$ n^2 -k^2= \pm 10 $$

$$(n-k)(n+k)=\pm 10 $$

$$(n-k)(n+k)= (\pm 2)\times (\pm 5) $$

or $$(n-k)(n+k)=(\pm1)\times( \pm 10) $$ Which has two integral solution of $\{1,3\}$

Thus all integers $$n \in \{1, …, 100 \}$$ except $\{1,3\}$ are in the desired set.

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