Find all pairs (p,q) such that the equation x4+2px2+qx+p2−36=0has exactly 4 integer roots(counting multiplicity).
Problem
Source: 2019 FKMO Problem 5
Tags: algebra, polynomial
24.03.2019 08:39
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
24.03.2019 09:35
This was like an calculus problem; my solution involves Rolle's theorem and some caseworks.
24.03.2019 09:38
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.
24.03.2019 10:45
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.
24.03.2019 11:16
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.