Mongolian Mathematical Olympiad

From the contrary, suppose there exist integers $(a,b)$ with $\gcd (a,b)=1$ and $|a|>|b|>1$ that satisfy the equation $$(a^n-b^n{)}^2=a^{n+m}-b^{n+m}$$ for $n\ge m\ge 1$. We can rewrite the given equation as follows: $$(a^n-b^n{)}^2=a^n(a^m-b^m)+b^m(a^n-b^n).$$ This implies that $a^n-b^n$ divides $a^m-b^m$. Let $S=\frac{a^m-b^m}{a^n-b^n}$. Then, dividing the equation by $a^n-b^n$, we have: $$a^n-b^n=a^{n}S+b^m,$$ that simplifies to: $$a^n(1-S)=b^m(b^{n-m}+1).$$ Since $|b|>1$ and $b^{n-m}+1\ne 0$, we conclude that $a^n$ divides $b^{n-m}+1$. Thus, $$a^n\mid b^{n-m}+1.$$ Now, consider the inequality: $$|b{|}^n+1\le (|b|+1{)}^n\le |a{|}^n\le |b{|}^{n-m}+1.$$ This implies: $|b{|}^n+1<|b{|}^n+1$, which is a contradiction. Hence, there are no pairs $(a,b)$ that satisfy the conditions of the problem.