Japan Mathematical Olympiad

When the condition is satisfied, there exists a non-negative integer $n$ such that $a{p}^4+2p^3+2p^2+1=n^2$ and then $$(a-p^2-2p-1)p^4=n^2-(p^3+p^2+1{)}^2=(n+p^3+p^2+1)(n-p^3-p^2-1)(\ast )$$ holds. If we assume that both $n+p^3+p^2+1$ and $n-p^3-p^2-1$ are divisible by $p$, then $n+1\equiv 0\equiv n-1(\mathrm{mod}p)$, leading to $2\equiv 0(\mathrm{mod}p)$, which contradicts the assumption that $p$ is an odd prime. Therefore, at least one of $n+p^3+p^2+1,n-p^3-p^2-1$ is not divisible by $p$. Since the left hand side of () is divisible by $p^4$, one of $n+p^3+p^2+1,n-p^3-p^2-1$ is divisible by $p^4$. In the case $n+p^3+p^2+1$ is divisible by $p^4$, there exists a positive integer $k$ such that $n=k{p}^4-p^3-p^2-1$. By substituting this into (), we obtain $(a-p^2-2p-1)p^4=k{p}^4(k{p}^4-2p^3-2p^2-2)$, hence $a=p^2+2p+1+k(k{p}^4-2p^3-2p^2-2)$. If $k\ge 2$, we have $$a>4(p^4-p^3-p^2-1)=4p^4\left(1-\frac{1}{p}-\frac{1}{p^2}-\frac{1}{p^4}\right)\ge 4p^4\left(1-\frac{1}{3}-\frac{1}{9}-\frac{1}{81}\right)>p^4$$ which contradicts $a<p^4$. If $k=1$, then $a=p^4-2p^3-p^2+2p-1=(p^2-p-1{)}^2-2$. In this case, by $p\ge 3$ we get $2\le p^2-p-1<p^2$ and then $1\le a<p^4$. Therefore, the number of pairs $(p,a)$ satisfying the condition is equal to the number of odd prime $p$ such that $(p^2-p-1{)}^2-2\le 2024$, and there exist three such primes $p=3,5,7$. In the case $n-p^3-p^2-1$ is divisible by $p^4$, there exists a non-negative integer $k$ such that $n=k{p}^4+p^3+p^2+1$. By substituting this to (*), we obtain $(a-p^2-2p-1)p^4=k{p}^4(k{p}^4+2p^3+2p^2+2)$, hence $a=p^2+2p+1+k(k{p}^4+2p^3+2p^2+2)$. If $k\ge 1$, then we have $a>p^4$ which contradicts $a<p^4$. If $k=0$, then we have $a=p^2+2p+1=(p+1{)}^2$. In this case, by $p\ge 3$ we get $1<(p+1{)}^2<(p^2{)}^2=p^4$ and then $1\le a<p^4$. Therefore, the number of pairs $(p,a)$ satisfying the condition is equal to the number of odd prime $p$ such that $(p+1{)}^2\le 2024$, and there exist 13 such primes $p=3,5,7,11,13,17,19,23,29,31,37,41,43$. We have proved that the total number of pairs $(p,a)$ satisfying the condition is $3+13=16$.