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The equation is $\displaystyle\frac{k(2-k)}{\sqrt{4k-3}}$. I have determined that it's integer when $k\in \{2,3,7,57\}$. Are there any other possibilities? How do I find all of them (and convince myself that I've found them all)?

Thank you!

Fabrizio Gambelín
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Adam Kennedy
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    It looks like the function is continuous from $(0,-\infty)$ so it is integral at infinitely many points. But @player3236 has the case for $k \in \mathbb{Z}$. – Derek Luna Nov 11 '20 at 06:46

1 Answers1

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Assuming you want $k \in \mathbb N$, you missed the case $k=1$.

Let $x =\dfrac {k(2-k)}{\sqrt{4k-3}}$. Then if $x \in \mathbb Z$, $256 x^2 \in \mathbb Z$.

We have:

$$256x^2 = \frac {256 (2 - k)^2 k^2}{4 k - 3} = 64 k^3 - 208 k^2 + 100 k + \frac {225}{4 k - 3} + 75$$

so if $x\ne 0$, $4k-3$ must be a divisor of $225$, and since $\sqrt{4k-3}$ is in the denominator, $4k-3$ must also be square. This gives the cases:

$$(k=2) \text{ or } (4k-3 = 1,9,25,225)$$

player3236
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