How many fish could there be?

Algebra Level 3

Andrew, a professional mathematician, goes to his local aquarium to buy two fish. He is told the number of female fish and male fish by the owner. While choosing the fish, Andrew notes that, if one was to choose two fish randomly, the chance of getting two fish of the same sex would be the same as the chance of getting two fish of the opposite sex.

There are $n$ fish inside the aquarium, where $n$ is a positive integer and $2\leq n\leq 2018$ . For how many values of $n$ is Andrew's observation possible?

The answer is 43.

1 solution

Mark Hennings
Apr 10, 2018

If there are $m$ males and $f$ females in the tank, then we need $\frac{\binom{m}{2} + \binom{f}{2}}{\binom{m+f}{2}} \; = \; \frac{\binom{m}{1}\times \binom{f}{1}}{\binom{m+f}{2}}$ and hence $\begin{aligned} \binom{m}{2} + \binom{f}{2} & = \; mf \\ m(m-1) + f(f-1) & = \; 2mf \\ (m-f)^2 & = \; m+f \end{aligned}$ and hence $n = m+f = (m-f)^2$ must be a perfect square. If $n = N^2$ for some integer $N \ge 2$ , then we have $\{m,f\} \,=\, \{\tfrac12N(N-1),\tfrac12N(N+1)\}$ . Thus the number of possible values of $n$ is the number of perfect squares greater than $1$ and less than or equal to $2018$ . Since $\sqrt{2018} = 44.922...$ , we deduce that the answer is $\boxed{43}$ .

How many fish could there be?

The answer is 43.

1 solution

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