MMO2025 Round 2

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 $\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 25,\pm 27,\pm 45,\pm 75,\pm 81,\pm 135,\pm 225,\pm 405,\pm 675,\pm 2025$.

Check each possible $a$:

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

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

Expand the product:

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

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

- $X^5$: $1$ - $X^4$: $a+c=0$ $⟹$ $c=-a$ - $X^3$: $ac+b+d=0$ - $X^2$: $ad+bc+e=0$ - $X^1$: $ae+bd=61$ - 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 $|b|,|e|\le 2025$.

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

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

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

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

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

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

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.