Prove that there are no positive integers $a, b$ and $n >1$ such that $a^n – b^n$ divides $ a^n + b^n$.

Question

Prove that there are no positive integers $a$ , $b$ and $n>1$ such that $a^{n}–b^{n}$ divides $a^{n}+b^{n}$.

Can someone provide me a proof of this and explain it to me please.

Answer 1

This is a problem that appears in this number theory book by Nivens et al (1.2.46, p 19). The following is a solution readily attained.

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Problem: Prove that there are no positive integers $a,b,n > 1$ such that $(a^n-b^n)\mid (a^n+b^n).$

Proof. We may first prove the following fact: $$ \text{For the integers $x,y$ such that $(x,y)=1, (x+y,x-y)=1$ or $2$}. $$ Since $(x,y)=1$, there is an integer $s,t$ such that $sx+ty=1$. Then $$ (s+t)(x+y)+(s-t)(x-y)=2sx+2ty=2. $$ Hence, $(x+y,x-y)\leq 2$ (by elementary number theory facts concerning $\mathrm{gcd}$); thus, we just proved the fact above.

Now, suppose that there are positive integers $a,b$ and $n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. Note that we may assume $a>b$ without loss of generality. Then there is a positive integer $Q$ which satisfies $$ (a^n+b^n)\mid Q(a^n-b^n). $$ Let $(a,b)=d$. Then divide this equation by $d^n$ to get that $$ \left(\left(\frac{a}{d}\right)^n-\left(\frac{b}{d}\right)^n\right) = Q\left(\left(\frac{a}{d}\right)^n+\left(\frac{b}{d}\right)^n\right) $$ and $\left(\left(\frac{a}{d}\right),\left(\frac{b}{d}\right)\right)=1$. Therefore, we can assume that there are positive integers $a,b$ which are relatively prime with $a>b$ and $n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. Now, $(a,b)=1$ implies that $(a^n,b^n)=1$, so we get that $$ a^n-b^n = (a^n-b^n,a^n+b^n)=1\;\text{or}\;2, $$ where first equality holds by $(a^n-b^n)\mid (a^n+b^n)$ and second equality holds by the fact we first proved.

But $a^n-b^n=(a-b)(a^{n-1}+\cdot+b^{n-1})$ and it is trivial that $a\neq b$ to make $(a^n-b^n)\mid (a^n+b^n)$ sense. So, $a>b\geq 1$. Therefore, $(a-b)\geq 1$ and $a^{n-1}+\cdots+b^{n-1}\geq 2+1=3$, since $n > 1$. Then we get that $$ a^n-b^n=(a-b)(a^{n-1}+\cdots+b^{n-1}\geq 3, $$ a contradiction, because it cannot be $1$ or $2$. Therefore, there are no positive integers $a,b,n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. $\Box$