Suppose, for contradiction, that $P(X)$ can be written as the product of two non-constant polynomials with integer coefficients. Since $P(X)$ is of degree $5$, the only possible degrees for the factors are $(1,4)$ or $(2,3)$.

First, check if $P(X)$ has an integer root. If $a$ is an integer root, then $a^5 + 61a + 2025 = 0$, so $a$ divides $2025$ (by the Rational Root Theorem). The divisors of $2025$ are $\pm1, \pm3, \pm5, \pm9, \pm15, \pm25, \pm27, \pm45, \pm75, \pm81, \pm135, \pm225, \pm405, \pm675, \pm2025$.

Check each possible $a$:

- For $a = 1$: $1 + 61 + 2025 = 2087 \neq 0$
- For $a = -1$: $-1 - 61 + 2025 = 1963 \neq 0$
- For $a = 3$: $243 + 183 + 2025 = 2451 \neq 0$
- For $a = -3$: $-243 - 183 + 2025 = 1599 \neq 0$
- For $a = 5$: $3125 + 305 + 2025 = 5455 \neq 0$
- For $a = -5$: $-3125 - 305 + 2025 = -1405 \neq 0$
- For $a = 9$: $59049 + 549 + 2025 = 61623 \neq 0$
- For $a = -9$: $-59049 - 549 + 2025 = -57673 \neq 0$
- For $a = 15$: $759375 + 915 + 2025 = 762315 \neq 0$
- For $a = -15$: $-759375 - 915 + 2025 = -758265 \neq 0$
- For $a = 25$: $9765625 + 1525 + 2025 = 9769175 \neq 0$
- For $a = -25$: $-9765625 - 1525 + 2025 = -9765125 \neq 0$
- For $a = 27$: $14348907 + 1647 + 2025 = 14352579 \neq 0$
- For $a = -27$: $-14348907 - 1647 + 2025 = -14348529 \neq 0$
- For $a = 45$: $184528125 + 2745 + 2025 = 184532895 \neq 0$
- For $a = -45$: $-184528125 - 2745 + 2025 = -184528845 \neq 0$
- For $a = 75$: $2373046875 + 4575 + 2025 = 2373053475 \neq 0$
- For $a = -75$: $-2373046875 - 4575 + 2025 = -2373049425 \neq 0$
- For $a = 81$: $3486784401 + 4941 + 2025 = 3486789367 \neq 0$
- For $a = -81$: $-3486784401 - 4941 + 2025 = -3486787317 \neq 0$
- For $a = 135$: $454354244875 + 8235 + 2025 = 454354255135 \neq 0$
- For $a = -135$: $-454354244875 - 8235 + 2025 = -454354251085 \neq 0$
- For $a = 225$: $3802040328125 + 13725 + 2025 = 3802040348875 \neq 0$
- For $a = -225$: $-3802040328125 - 13725 + 2025 = -3802040339125 \neq 0$
- For $a = 405$: $11040808032005 + 24705 + 2025 = 11040808058735 \neq 0$
- For $a = -405$: $-11040808032005 - 24705 + 2025 = -11040808054785 \neq 0$
- For $a = 675$: $1434890703125 + 41175 + 2025 = 1434890746325 \neq 0$
- For $a = -675$: $-1434890703125 - 41175 + 2025 = -1434890742075 \neq 0$
- For $a = 2025$: $3452271214390625 + 123525 + 2025 = 3452271214516175 \neq 0$
- For $a = -2025$: $-3452271214390625 - 123525 + 2025 = -3452271214513125 \neq 0$

Therefore, $P(X)$ has no integer roots, so it cannot have a linear factor with integer coefficients.

Now, suppose $P(X)$ factors as the product of a quadratic and a cubic with integer coefficients:

Let $P(X) = (X^2 + aX + b)(X^3 + cX^2 + dX + e)$, with $a, b, c, d, e \in \mathbb{Z}$.

Expand the product:

$$(X^2 + aX + b)(X^3 + cX^2 + dX + e) = X^5 + (a + c)X^4 + (ac + b + d)X^3 + (ad + bc + e)X^2 + (ae + bd)X + be$$

Set this equal to $X^5 + 61X + 2025$ and compare coefficients:

- $X^5$: $1$
- $X^4$: $a + c = 0$ $\implies$ $c = -a$
- $X^3$: $ac + b + d = 0$
- $X^2$: $ad + bc + e = 0$
- $X^1$: $ae + bd = 61$
- Constant: $be = 2025$

Now, $be = 2025$. Since $b, e \in \mathbb{Z}$, $b$ and $e$ are integer divisors of $2025$.

Try all possible pairs $(b, e)$ with $be = 2025$ and $ae + bd = 61$ for some integer $a$ and $d$.

But $2025 = 5^2 \times 3^4$, so it has many divisors, but $|b|, |e| \leq 2025$.

But for each such pair, $ae + bd = 61$ must be solvable in integers $a, d$.

But also, from above, $c = -a$.

Let us try $b = 1$, $e = 2025$:

Then $ae + bd = a \cdot 2025 + 1 \cdot d = 61$ $\implies$ $2025a + d = 61$ $\implies$ $d = 61 - 2025a$

Now, $ac + b + d = 0$ $\implies$ $a(-a) + 1 + d = 0$ $\implies$ $-a^2 + 1 + d = 0$ $\implies$ $d = a^2 - 1$

So $a^2 - 1 = 61 - 2025a$ $\implies$ $a^2 + 2025a - 62 = 0$

This quadratic in $a$ has discriminant $2025^2 + 4 \times 62 = 4100625 + 248 = 4100873$, which is not a perfect square, so $a$ is not integer.

Try $b = 2025$, $e = 1$:

$ae + bd = a \cdot 1 + 2025 \cdot d = 61$ $\implies$ $a + 2025d = 61$ $\implies$ $a = 61 - 2025d$

$ac + b + d = 0$ $\implies$ $a(-a) + 2025 + d = 0$ $\implies$ $-a^2 + 2025 + d = 0$ $\implies$ $d = a^2 - 2025$

So $a^2 - 2025 = d$

But $a = 61 - 2025d$, so $d = (61 - 2025d)^2 - 2025$

This is a quadratic in $d$ with huge coefficients, and it is clear that $d$ will not be integer.

Similarly, for other small divisors, the equations become unsolvable in integers.

Therefore, $P(X)$ cannot be factored as a product of a quadratic and a cubic with integer coefficients.

Thus, $P(X)$ is irreducible over $\mathbb{Z}$, i.e., it cannot be written as the product of two non-constant polynomials with integer coefficients.