Let p be a certain prime number. Find all non-negative integers n for which polynomial P(x)=x4-2(n+p)x2+(n-p)2 may be rewritten as product of two quadratic polynomials with integer coefficients.
let
(x^2+ax+b)(x^2+cx+d)=P(x)
then we have
bd=(n-p)^2
ad=-bc
ac+b+d=-2(n+p)
a=-c -> c=-a
substituting gives
bd=(n-p)^2
ad=ab
-a^2+b+d=-2(n+p)
case 1: a=0
bd=(n-p)^2
b+d=-2(n+p)
d=-2(n+p)-b
b[-2(n+p)-b]=(n-p)^2
b^2+2(n+p)b+(n-p)^2=0
b=-(sqrt(n)+-sqrt(p))^2
n=p
case 2: a!=0
d=b
b^2=(n-p)^2
-a^2+2b=-2(n+p)
b=+-(n-p)
sub case 1: b=-(n-p)=p-n
-a^2+2p-2n=-2n-2p
-a^2=-4p
a^2=4p false
sub case 2: b=n-p
-a^2+2n-2p=-2n-2p
-a^2=-4n
a^2=4n
n=k^2
a^2=4k^2
a=2k
b=2k-p
c=-2k
d=2k-p
(x^2+2k+2k-p)(x^2-2k+2k-p)
so the possible values for n are either p or any perfect square
|
|
Posted by Daniel
on 2020-12-29 06:20:47 |
solution
