MMO2025 Round 4

Answer: $(m,n,p)=(1,1,2)$.

Setting $k=2024$, $m_1=m/d$ and $n_1=n/d$, the given identity becomes $d^{k-2}(d{m}_1^{k+1}+n_1^k)=p{m}_1{n}_1$, where $d$ is the greatest common divisor $(m,n)$ of $n$ and $m$. Put $a=n_1/b$ and $c=d/b$, where $b$ denotes $(d,n_1)$. Then we get $(cb{)}^{k-2}(c{m}_1^{k+1}+a^k{b}^{k-1})=p{m}_{1}a$. Observe that $c{m}_1^{k+1}+a^k{b}^{k-1}$ is greater than 1 and is relatively prime to $m_{1}a$. Then $(cb{)}^{k-2}$ is divisible by $m_{1}a$ and $$\frac{(cb{)}^{k-2}}{m_{1}a}(c{m}_1^{k+1}+a^k{b}^{k-1})=p(0.3)$$ which implies $(cb{)}^{k-2}=m_{1}a$. Since $(m_1,b)=1$ and $(a,c)=1$ we must have $a=b^{k-2}$ and $m_1=c^{k-2}$. Thus, it follows from (0.3) that $c^{k^2-k+1}+b^{k^2-k+1}=p$, hence $p$ must be divisible by $b+c$ since $k^2-k+1$ is odd. Therefore $p=c+b$ which gives us $c+b=c^{k^2-k+1}+b^{k^2-k+1}$. Hence we conclude that $c=b=1$ and so $m=n=1,p=2$.