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.

I am writing this answer because you said that you were trying a guess and check method. Computers are good at this. A decent algorithm is to have two integers $n_x$ and $n_y$ which start at 1. Then, calculate x by doing $2n_x^2$ and y by doing $5n_y^5$. Check if they are equal; if they are, you found your answer. If not, whichever of $x$ and $y$ are lower, increment that $n$ value (i.e., if $x < y$, then increment $n_x$). Recalculate $x$ and $y$ and repeat until they are the same.

Here is an example implementation in Python using generators:

class SpecialSquareGenerator:

    def __init__(self, n=0):
        self.n = n

    def __iter__(self):
        return self

    def __next__(self):
        self.n += 1
        return self.n, 2*(self.n**2)

class SpecialFifthGenerator:

    def __init__(self, n=0):
        self.n = n

    def __iter__(self):
        return self

    def __next__(self):
        self.n += 1
        return self.n, 5*(self.n**5)

def special_square():
    n = 0;
    ss = SpecialSquareGenerator()
    sf = SpecialFifthGenerator()
    nx, x = next(ss)
    ny, y = next(sf)
    print("{0}: {1}\t{2}: {3}".format(nx, x, ny, y))
    while True:
        if (x == y): return x
        if x < y:
            nx, x = next(ss)
        else:
            ny, y = next(sf)
        print("{0}: {1}\t{2}: {3}".format(nx, x, ny, y))

if __name__ == "__main__":
    print(special_square())

Running it returns the right answer:

gns-mac1:sandbox gns$ python3 special_square.py 
1: 2    1: 5
2: 8    1: 5
2: 8    2: 160
3: 18   2: 160
...(output omitted)
494: 488072 10: 500000
495: 490050 10: 500000
496: 492032 10: 500000
497: 494018 10: 500000
498: 496008 10: 500000
499: 498002 10: 500000
500: 500000 10: 500000
500000

Of course, the mathematical approach is better for understanding the problem. But if you need to guess and check, then computers are the tool for that.

P.S.

There is another way to exhaustively search for the solution. You can take sequential numbers and try dividing them by 2 (or 5) and then taking the square root (or fifth root) and then checking if that result is an integer for both operations. There are two downsides to this approach:

• You have to decide if a floating point number is supposed to represent an integer. This is hard for computers and language implementations to do because computers only have a fixed set of digits to represent floating point numbers.
• The search space is greater (by order of $n^2$). So that means that you should expect to take longer to reach the same answer, given the same hardware.

P.S.S.

There are faster ways to implement both my algorithm, and the other I mentioned in the postscript. For example, you can double $n$ each time and then when you overshoot, use binary search in the space between the last $n$ and the one that overshot.