What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?

The number is clearly a multiple of $5$ and $2$. We look for the smallest, so we assume that it has no more prime factors.

So let $n=2^a5^b$. Since $n/2$ is a square, then $a-1$ and $b$ are even. Since $n/5$ is a fifth power, $a$ and $b-1$ are multiples of $5$. Then $a=5$ and $b=6$.

Here's a very unsophisticated approach: Let $n$ be the smallest such integer. Then there exist integers $a$ and $b$ such that $n=5a^5$ and $n=2b^2$. It follows that $a$ is a multiple of $2$, say $a=2a_1$, and $b$ is a multiple of $5$, say $b=5b_1$. Then $$n=2^5\cdot5\cdot a_1^5\qquad\text{ and }\qquad n=2\cdot5^2\cdot b_1^2.$$ This in turn shows that $a_1$ is a multiple of $5$, say $a_1=5a_2$, and $b_1$ is a multiple of $2$, say $b_1=2b_2$. Then $$n=2^5\cdot5^6\cdot a_2^5\qquad\text{ and }\qquad n=2^3\cdot5^2\cdot b_2^2.$$ This in turn shows that $b_2$ is a multiple of both $2$ and $5^2$, say $b_2=2\cdot5^2\cdot b_3$. Then $$n=2^5\cdot5^6\cdot a_2^5\qquad\text{ and }\qquad n=2^5\cdot5^6\cdot b_3^2.$$ This shows that $n\geq2^5\cdot5^6$, and as you might expect a quick check shows that $n=2^5\cdot5^6$ does indeed work, so $n=2^5\cdot5^6=500000$.

This is like code golf...

The answer is 500000.

Proof by computation: (in R)

> x=(1:10)^5*5
> x
 [1]      5    160   1215   5120  15625  38880  84035 163840
 [9] 295245 500000
> sqrt(x/2)
 [1]   1.581139   8.944272  24.647515  50.596443  88.388348
 [6] 139.427400 204.981707 286.216701 384.216736 500.000000

Done.