# 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 $k^4$

• 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. – Arief Anbiya Apr 1 '18 at 9:40
• @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. – Prathyush Poduval Apr 1 '18 at 9:41
• @arief this is exactly the contradiction. You get $n+k$ is even because $n-k$ is even and also $n+k$ odd since $2||10$. – Or Kedar Apr 1 '18 at 9:48
• @PrathyushPoduval Understand.. – Arief Anbiya Apr 1 '18 at 10:03

$$n^{4} - 20n^{2} + 100=$$

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

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

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

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

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.