Dana Heuberger has posted an unusual problem and its solution at the CutTheKnotMath facebook page.

Let $a\in\mathbb{R},\,k,n\in\mathbb{N}^{*}=\{x\in\mathbb{N}:\,x\ge 1\},$ with $\gcd(n,k)=1.$ Consider the equation

(1)

$\{a^k\}x^2-4\{a^n\}x+6=0.$

Assume (1) has a (real) solution. Prove then that

1. if the solutions of (1) are integer, then $a\in\mathbb{Q}.$

2. for $n\in\mathbb{N}^{*},$ there are infinitely many values of $a\gt 1,$ such that the solutions of equation (1), with $k=2n+1,$ are integer.

(Here $\{\alpha\}=\alpha-\lfloor\alpha\rfloor,$ the fractional part of $\alpha .)$

### Proof

For the first part, we shall consider three cases:

1. $a^{k},a^{n}\in\mathbb{Z}$ leads to $6=0$ which is false.

2. $a^{k}\in\mathbb{Z},\,a^{n}\not\in\mathbb{Z}.$ The equation then has the solution $\displaystyle x=\frac{3}{2\{a^n\}}\in\mathbb{Z},$ implying $a^n\in\mathbb{Q}.$

Let $p,q\in\mathbb{Z}$ be such that $np+kq=1.$ Then

$a=(a^n)^q\cdot (a^k)^p\in\mathbb{Q}.$

3. If $a^k\not\in\mathbb{Z},$ equation (1) has two integer roots $x_1,\,x_2\in\mathbb{Z}.$

We have $\displaystyle x_1x_2=\frac{6}{\{a^k\}},$ so that $\displaystyle \{a^k\}=\frac{6}{x_1x_2}\in\mathbb{Q}.$ But then $\displaystyle x_1+x_2=\frac{4\{a^n\}}{\{a^k\}}=\frac{2x_1x_2}{3}\{a^n\}\in\mathbb{N}$ and $\{a^n\}\in\mathbb{Q}$ such that $a^n\in\mathbb{Q}.$ As in the previous case this implies $a\in\mathbb{Q}.$

For the second part, observe that, for $\displaystyle a=\frac{1}{2},$ the equation becomes $\displaystyle\frac{x^2}{2^{2n+1}}-\frac{4x}{2^n}+6=0,$ i.e., $x^2-2^{n+3}x+3\cdot 2^{2n+2}=0,$ with two solutions $x_1=2^{n+1}$ and $x_2=3\cdot 2^{n+1}.$

Thus, for all $s\in\mathbb{N}^{*},$ $\displaystyle a_s=2^{2n}\cdot s+\frac{1}{2}$ answers the second part because by Newton's binomial theorem, $\displaystyle\{a_{s}^{2n+1}\}=\frac{1}{2^{2n+1}}$ and $\displaystyle\{a_{s}^{n}\}=\frac{1}{2^{n}}$ which give the same equation as before, with both solutions integer.