Subfield: Number Theory


Question: Prove that $5^{n}-3^{n}$ is not divisible by $2^{n}+65$ for any positive integer $n$.

Solution:
Let $n$ be a positive integer, and let $m=2^{n}+65$. For the sake of contradiction, suppose that $m \mid 5^{n}-3^{n}$, so $5^{n} \equiv 3^{n}(\bmod m)$.

Notice that if $n$ is even, then $3 \mid m$, but $3 \nmid 5^{n}-3^{n}$, contradiction. So, from now on we assume that $n$ is odd, $n=2 k+1$. Obviously $n=1$ is not possible, so $n \geqslant 3$. Notice that $m$ is coprime to 2,3 and 5 .

Let $m_{1}$ be the smallest positive multiple of $m$ that can be written in the form of either $\left|5 a^{2}-3 b^{2}\right|$ or $\left|a^{2}-15 b^{2}\right|$ with some integers $a$ and $b$.

Note that $5^{n}-3^{n}=5\left(5^{k}\right)^{2}-3\left(3^{k}\right)^{2}$ is a multiple of $m$, so the set of such multiples is non-empty, and therefore $m_{1}$ is well-defined.

I. First we show that $m_{1} \leqslant 5 m$. Consider the numbers
$$
5^{k+1} x+3^{k+1} y, \quad 0 \leqslant x, y \leqslant \sqrt{m}
$$
There are $|\sqrt{m}|+1>\sqrt{m}$ choices for $x$ and $y$, so there are more than $m$ possible pairs $(x, y)$. Hence, two of these sums are congruent modulo $m$ : $5^{k+1} x_{1}+3^{k+1} y_{1} \equiv 5^{k+1} x_{2}+3^{k+1} y_{2}(\bmod m)$.

Now choose $a=x_{1}-x_{2}$ and $b=y_{1}-y_{2}$; at least one of $a, b$ is nonzero, and
$$
5^{k+1} a+3^{k+1} b \equiv 0 \quad(\bmod m), \quad|a|,|b| \leqslant \sqrt{m}
$$
From
$$
0 \equiv\left(5^{k+1} a\right)^{2}-\left(3^{k+1} b\right)^{2}=5^{n+1} a^{2}-3^{n+1} b^{2} \equiv 5 \cdot 3^{n} a^{2}-3^{n+1} b^{2}=3^{n}\left(5 a^{2}-3 b^{2}\right) \quad(\bmod m)
$$
we can see that $\left|5 a^{2}-3 b^{2}\right|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5 a^{2} \neq 3 b^{2}$. Hence, by the choice of $a, b$, we have $0<\left|5 a^{2}-3 b^{2}\right| \leqslant \max \left(5 a^{2}, 3 b^{2}\right) \leqslant 5 m$. That shows that $m_{1} \leqslant 5 m$

II. Next, we show that $m_{1}$ cannot be divisible by 2,3 and 5 . Since $m_{1}$ equals either $\left|5 a^{2}-3 b^{2}\right|$ or $\left|a^{2}-15 b^{2}\right|$ with some integers $a, b$, we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of $m$, smaller than $m_{1}$.

- If $5 \mid m_{1}$ and $m_{1}=\left|5 a^{2}-3 b^{2}\right|$, then $5 \mid b$ and $\left|a^{2}-15\left(\frac{b}{5}\right)^{2}\right|=\frac{m_{1}}{5}<m_{1}$.
- If $5 \mid m_{1}$ and $m_{1}=\left|a^{2}-15 b^{2}\right|$, then $5 \mid a$ and $\left|5\left(\frac{a}{5}\right)^{2}-3 b^{2}\right|=\frac{m_{1}}{5}<m_{1}$.
- If $3 \mid m_{1}$ and $m_{1}=\left|5 a^{2}-3 b^{2}\right|$, then $3 \mid a$ and $\left|b^{2}-15\left(\frac{a}{3}\right)^{2}\right|=\frac{m_{1}}{3}<m_{1}$.
- If $3 \mid m_{1}$ and $m_{1}=\left|a^{2}-15 b^{2}\right|$, then $3 \mid a$ and $\left|5 b^{2}-3\left(\frac{a}{3}\right)^{2}\right|=\frac{m_{1}}{3}<m_{1}$.
- If $2 \mid m_{1}$ and $m_{1}=\left|5 a^{2}-3 b^{2}\right|$, then $\left|\left(\frac{5 a-3 b}{2}\right)^{2}-15\left(\frac{a-b}{2}\right)^{2}\right|=\frac{m_{1}}{2}<m_{1}$.
- If $2 \mid m_{1}$ and $m_{1}=\left|a^{2}-15 b^{2}\right|$, then $\left|5\left(\frac{a-3 b}{2}\right)^{2}-3\left(\frac{a-5 b}{2}\right)^{2}\right|=\frac{m_{1}}{2}<m_{1}$.

(The last two expressions can be obtained from $(\sqrt{5} a+\sqrt{3} b)(\sqrt{5}-\sqrt{3})=(5 a-3 b)+\sqrt{15}(b-a)$ and $(a+\sqrt{15} b)(\sqrt{5}-\sqrt{3})=\sqrt{5}(a-3 b)+\sqrt{3}(5 b-a)$.)

In all six cases, we found that either $\frac{m_{1}}{2}, \frac{m_{1}}{3}$ or $\frac{m_{1}}{5}$ is of the form $\left|5 x^{2}-3 y^{2}\right|$ or $\left|x^{2}-15 y^{2}\right|$. Since $m$ is coprime to 2,3 and 5 , the presented number is a multiple of $m$, but this contradicts the minimality of $m_{1}$.

III. The last remaining case is $m_{1}=m$, so either $m=\left|5 a^{2}-3 b^{2}\right|$ or $m=\left|a^{2}-15 b^{2}\right|$. We will get a contradiction by considering the two sides modulo 3,4 and 5 .



- $2^{n}+65=5 a^{2}-3 b^{2}$ is not possible, because $2^{n}+65 \equiv 1(\bmod 3)$, but $5 a^{2}-3 b^{2} \not \equiv 1$ $(\bmod 3)$.
- $2^{n}+65=3 b^{2}-5 a^{2}$ is not possible, because $2^{n}+65 \equiv 1(\bmod 4)$, but $3 b^{2}-5 a^{2} \not \equiv 1$ $(\bmod 4)$.
- $2^{n}+65=a^{2}-15 b^{2}$ is not possible, because $2^{n}+65 \equiv \pm 2(\bmod 5)$, but $a^{2}-15 b^{2} \not \equiv \pm 2$ $(\bmod 5)$.
- $2^{n}+65=15 b^{2}-a^{2}$ is not possible, because $2^{n}+65 \equiv 1(\bmod 4)$, but $15 b^{2}-a^{2} \not \equiv 1$ $(\bmod 4)$.

We found a contradiction in all cases, that completes the solution.
Notice that if $n$ is even, then $3 \mid m$, but $3 \nmid 5^{n}-3^{n}$, contradiction. So, from now on we assume that $n$ is odd, $n=2 k+1$. Obviously $n=1$ is not possible, so $n \geqslant 3$. Notice that $m$ is coprime to 2,3 and 5 .

Suppose again that $5^{n} \equiv 3^{n}\left(\bmod m=2^{n}+65\right)$. Like above, we conclude that $n$ must be odd, and $n \geqslant 3$, so $8 \mid 2^{n}$.

Using Jacobi symbols,
$$
-1=\left(\frac{2^{n}+65}{5}\right)=\left(\frac{5}{2^{n}+65}\right)=\left(\frac{5^{n}}{2^{n}+65}\right)=\left(\frac{3^{n}}{2^{n}+65}\right)=\left(\frac{3}{2^{n}+65}\right)=\left(\frac{2^{n}+65}{3}\right)=1,
$$

contradiction.