Exercise 1.2.46* ($a^n - b^n$ doesn't divide $a^n + b^n$)

Proof. Let $n>1$. Reasoning by contradiction assume that there are positive integers $a,b$ such that $a^n-b^n\mid a^n+b^n$.

(a) First reduction. We will prove that there are relatively prime positive integers $a^{\prime },b^{\prime }$ such that ${a{}^{\prime }}^n-{b{}^{\prime }}^n\mid {a{}^{\prime }}^n+{b{}^{\prime }}^n$.

Write $a=d{a}^{\prime },b=d{b}^{\prime }$, where $d=a\wedge b>0$, and $a^{\prime }\wedge b^{\prime }=1$. Then $d\mid a,d\mid b$, so $d^n\mid a^n,d^n\mid b^n$.

Dividing by $d^n$, we obtain ${\left(\frac{a}{d}\right)}^n-{\left(\frac{b}{d}\right)}^n\mid {\left(\frac{a}{d}\right)}^n+{\left(\frac{b}{d}\right)}^n$, that is

Since ${b{}^{\prime }}^n-{a{}^{\prime }}^n=-({a{}^{\prime }}^n-{b{}^{\prime }}^n)\mid {b{}^{\prime }}^n+{a{}^{\prime }}^n$, possibly exchanging $a^{\prime },b^{\prime }$, we can assume that $b^{\prime }\ge a^{\prime }\ge 1$, and $a^{\prime }\ne b^{\prime }$, otherwise $0\mid {a{}^{\prime }}^n+{b{}^{\prime }}^n$.

Renaming $a^{\prime },b^{\prime }$, we conclude that there are integers $a,b$ such that

(b) Since $a^n-b^n\mid a^n+b^n$ and $a^n-b^n\mid a^n-b^n$, then

Moreover $a\wedge b=1$, thus $a^n\wedge b^n=1$ and $(a^n-b^n)\wedge a^n=1$. Since $a^n-b^n\mid 2a^n$, we obtain

where $a>b$. Therefore $a^n-b^n=1$ or $a^n-b^n=2$.

(c)

• If $a^n-b^n=1$, the formula $a^n-b^n=(a-b)\underset{k=0}{\overset{n-1}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}^k{b}^{n-1-k}$ shows that $a-b\mid a^n-b^n=1$, and $a-b>0$, so $a-b=1$.

  $$1=a^n-b^n=\underset{k=0}{\overset{n-1}{\sum }}{a}^k{b}^{n-1-k}\ge n,$$

  because $a^k{b}^{n-1-k}\ge 1$ for every index $k=0,\dots ,n-1$. This gives $n\le 1$. This is a contradiction, because $n>1$ by hypothesis.

• If $a^n-b^n=2$, then $a,b$ are of same parity. Since $a\wedge b=1$, they are not both even, so $a,b$ are both odd. Therefore $a-b\ne 1$, and $a-b\mid a^n-b^n=2(a-b>0)$, thus $a-b=2$. Then

  $$2=a^n-b^n=2\underset{k=0}{\overset{n-1}{\sum }}{a}^k{b}^{n-1-k},$$

  therefore $1=\underset{k=0}{\overset{n-1}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}^k{b}^{n-1-k}\ge n$: same contradiction.

Both cases lead to a contradiction. This proves that there are no positive integers $a,b$, and $n>1$ such that $a^n-b^n\mid a^n+b^n$. □