China Girls' Mathematical Olympiad

Assuming there are positive integers $m,n$ such that $m^{20}+11^n=k^2$ with $k\in \mathbb{Z}$, we then have $$11^n=k^2-m^{20}=(k-m^{10})(k+m^{10}),$$ which means that there are integers $\alpha ,\beta \ge 0$ such that $$\{\begin{array}{l}k-m^{10}=11^{\alpha },\\ k+m^{10}=11^{\beta }.\end{array}\begin{array}{l}\stackrel{◯}{1}\\ \stackrel{◯}{2}\end{array}$$ It is obvious that $\alpha <\beta$. Subtracting ① from ②, we have $$2m^{10}=11^{\alpha }(11^{\beta -\alpha }-1).$$ Let $m=11^{\gamma }{m}_1$, where $\gamma ,m_1\in {\mathbb{N}}^{\ast }$, $11\nmid m_1$. Then $$11^{10\gamma }\cdot 2m_1^{10}=11^{\alpha }(11^{\beta -\alpha }-1),$$ which implies $10\gamma =\alpha$ and $2m_1^{10}=11^{\beta -\alpha }-1$. By Fermat's Little Theorem, we have $m_1^{10}\equiv 1(\mathrm{mod}11)$, then $2m_1^{10}\equiv 2(\mathrm{mod}11)$. But $11^{\beta -\alpha }-1\equiv 10(\mathrm{mod}11)$, which is a contradiction. So the proved conclusion is that there are no positive integers $m,n$ such that $m^{20}+11^n$ is a square number. $\square$