4-degree polynomial has integer roots

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4-degree polynomial has integer roots

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Source: 2019 FKMO Problem 5

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lminsl

#1 h Mar 24, 2019, 5:33 AM • 2 Y

Y by Adventure10, Mango247

Find all pairs $(p,q)$ such that the equation $x^4+2px^2+qx+p^2-36=0$ has exactly $4$ integer roots(counting multiplicity).

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#2 h Mar 24, 2019, 5:39 AM • 1 Y

Y by Adventure10

The answer is $(p,q)=(-6,16) (-6,-16) (-10,0)$
Key observation is to prove that $144=(ab-cd)^2+(ac-bd)^2+(ad-bc)^2$ , which $a,b,c,d$ are four roots of the equation

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lminsl

#3 h Mar 24, 2019, 6:35 AM • 2 Y

Y by Adventure10, Mango247

This was like an calculus problem; my solution involves Rolle's theorem and some caseworks.

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#4 h Mar 24, 2019, 6:38 AM • 1 Y

Y by Adventure10

Let $a,b,c,d$ be the four roots of the equation that $a\equiv b\pmod{2}$ .
$a+b+c+d=0$ gives us $c\equiv d\pmod{2}$ . Also WLOG assume $a^2\geqslant b^2,c^2\geqslant d^2,$ and $ab\geqslant cd$ .
We have $ab+cd-(a+b)^2=2p$ and $abcd=p^2-36$ .
This gives $(ab+cd)^2-2(ab+cd)(a+b)^2+(a+b)^4-4abcd=144$ .
Note that $(ab+cd)^2-2(ab+cd)(a+b)^2+(a+b)^4-4abcd = (ab-cd)^2+(a+b)^2((a+b)^2-2(ab+cd))=(ab-cd)^2+(a+b)^2(a^2+b^2-2cd).$ From $(a+b)^2=(c+d)^2$ , we get $a^2+b^2-2cd=c^2+d^2-2ab\implies a^2+b^2-2cd=\frac{(a^2+b^2-2cd)+(c^2+d^2-2ab)}{2}=\frac{(a-b)^2+(c-d)^2}{2}.$ Hence, we get $144=(ab-cd)^2+(a+b)^2\frac{(a-b)^2+(c-d)^2}{2}\implies 288=2(ab-cd)^2+(a^2-b^2)^2+(c^2-d^2)^2.$ We have $4\mid a^2-b^2,c^2-d^2\implies 16\mid 2(ab-cd)^2\implies 4\mid ab-cd$ . So,
$18=2\left( \frac{ab-cd}{4}\right)^2 +\left( \frac{a^2-b^2}{4}\right)^2 +\left( \frac{c^2-d^2}{4}\right)^2.$ Clearly, the only non-negative integer solutions $(x,y,z)$ to $18=2x^2+y^2+z^2$ are $(0,3,3),(1,4,0),(1,0,4),(2,3,1),(2,1,3),(3,0,0).$ From here the rest is easy.

This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Mar 24, 2019, 6:40 AM

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#5 h Mar 24, 2019, 7:45 AM • 2 Y

Y by Adventure10, Mango247

My solution
Let the four roots be $a,b,c,d$ . $a+b+c+d=0$ , $ab+bc+cd+da+ac+bd=2p$ , $abcd=p^2-36$ .
Just substitute $d=-a-b-c$ , and we get $\left(-a^2-b^2-c^2-ab-bc-ca\right)=2p$ , $-abc(a+b+c)=p^2-36$ .

Now let $x=b+c$ , $y=c+a$ , $z=a+b$ .
Then we get $x^2+y^2+z^2=-4p$ , $2\left(y^2z^2+z^2x^2+x^2y^2\right)-\left(x^4+y^4+z^4\right)=36-p^2$
Add the square of the first equation to the second equation , then $y^2z^2+z^2x^2+x^2y^2=144$ .
After analyzing by $\mod 2$ , we get that $x,y,z$ are all even.

$\therefore$ $\left(\frac{y}{2}\right)^2\left(\frac{z}{2}\right)^2+\left(\frac{z}{2}\right)^2\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2\left(\frac{y}{2}\right)^2=9$ .

Now it's just a matter of casework.

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#6 h Mar 24, 2019, 8:16 AM • 2 Y

Y by Adventure10, Mango247

By the way, I would like to see solution from outline in #3, can you provide more details?

lminsl wrote:

This was like an calculus problem; my solution involves Rolle's theorem and some caseworks.

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