Mexican Math Olympiad Problem.

Suppose $n$ is an odd number, then all of it's divisors will be odd and $n=d_2^2+d_3^2$ will be even. Contradiction, $n$ is even. It's immediate that $d_2=2$ and $n=4+d_3^3$ Now $d_3|n$ and $d_3|d_3^3$ , so we must have $d_3|4$ and $d_3=4$ since $d_3>2$ . $n=2^2+4^3=68$ $n=68$ indeed satisfies the conditions of problem.

Let $n = p_{1}^{a_{1}} p_{2}^{a_{2}} p_{3}^{a_{3}} ...p_{k}^{a_{k}}$ for a positive integer $k$ and $a_{k} \in N \forall k \in N$ .

It is obvious that $d_{2} = p_{1}$ .

While $d_{3}$ can have two forms which is : $d_{3} = p_{2}$ if and only if $p_{1}^{2} > p_{2}$ . That is if $p_{1}^{2} | n$ . Otherwise $d_{3} = p_{1}^{2}$ .

(i) $d_{3} = p_{2}$

We have $n = p_{1}^{2} + p_{2}^{3}$ . Let $p_{1} = 2$ , $2 | n$ .

Therefore $n = p_{2}^{3} + 4 \equiv -1, 1 (mod 4)$ as $p_{2} \geq 3$ . Thus $2 \nmid n$ . Contradiction.

Now let $p_{1} \geq 3$ , thus $2 \nmid n$ .

We have, $n = p_{2}^{3} + 1 \equiv 0, 2 (mod 4)$ . Thus $2 | n$ . Contradiction.

(ii) $d_{3} = p_{1}^{2}$

Therefore, $n = p_{1}^{2} + p_{1}^{6}$

Let $p_{1} \geq 3$ . Thus $2 \nmid n$ . But $n = p_{1}^{2} + p_{1}^{6} \equiv 0 (mod 2)$ . Contradiction.

Let $p_{1} = 2$ ,

We have $n = 4 + 64 = 68$ with $2^{2} < 17$ . Therefore $n = 68$ satisfies the above equality.

Let $n$ be such an integer. Let $d=(d_2,d_3)$ . Note that $d\leq d_2$ and $d|n$ , so $d=1$ or $d=d_2$ . Moreover, $d_1|(n-d_2^2)=d_3^3$ . Since $d_2,d_3>1$ , we conclude $d_2,d_3$ are not coprime, so $d=d_2$ . Thus, $d_3=kd_2$ for some integer $k$ .

Then $k|n$ and $k\le d_3$ , so $k=1$ or $k=d_2$ . However, $d_2\neq d_3$ , so $k\neq 1$ . Therefore, $d_3=d_2^2$ .

Note that $d_2$ is the smallest prime factor of $n$ . If $d_2$ is odd, then $n=d_2^2+d_2^6$ is even, so $2|n$ , yielding a contradiction. Therefore, $d_2$ is even, so $d_2=2$ .

Thus, $n=2^2+2^6=68$ . Moreover, $68$ satisfies the given conditions since its smallest factors greater than $1$ are indeed $2$ and $4$ . Therefore, the answer is $\boxed{68}$ .