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.
(In reply to
Puzzle Answer by K Sengupta)
Let p(x) = (x^2+qx+r)(x^2+sx+t)
Comparing the coefficients of x^0, x^1, x^2, x^3, and x^4, we have:
rt=(n-p)^2
q=-s
qs+r+t =-2(n+p)
qt=-rs
Now, q=-s, gives s=-q.
Then this substitution gives:
rt=(n-p)^2
qt=qr
-q^2+r+t= -2(n+p)
CASE 1: q is NOT 0
--------------------------------
t=r
So, t=(n-p)^2, gives: r^2=(n-p)^2
=> r= +/- (n-p)
Case- 1(i) r= -(n-p)
----------------------------
r= -(n-p) = p-n
Then,
- q^2 + 2r = - 2n-2p
=>- q^2+2p-2n = - 2n-2p
=> q^2 = 4p
q=2Vp
Then, we have :
p(x) =(x^2 +2Vp*x + p-n)(x^2+sx+p-n)
= x^4+(2Vp - s)x^3 +{2(p-n)+2Vp)x^2+ {2(p-n)Vp+(p-n)s}x+(p-n)^2
The coefficients x^3, x^2, and x involve Vp which is NOT ALWAYS an integer for all integer values of p.
This is a contradiction.
Case-1(ii): r= n-p
----------------------------
Then, we have:
-q^2+2n-2p= - 2n-2p
=> q^2 = 4n
This is possible only if n =m^2, where m is an integer.
Then,
q= 2m
r=2m-p
s=2m
t= 2m-p
Then the required factorization of p(x) is:
(x^2+2mx+2m-p)(x^2-2mx+2m-p)
Thus, when n is a perfect square, the each of the 2 two quadratic factors have integer coefficients.
CASE II: q = 0
-----------------------
rt = (n-p)^2
r+t = -2(n+p)
=> t = -2(n+p) - r
Therefore, it follows that:
r[-2(n+p)-r] = (n-p)^2
=> r^2+2(n+p )r+(n-p)^2=0
=> r = -(n+p) +/- 2*V(np) = -(Vn +/-Vp)^2
If, r= -(Vn -Vp), then for n=p, we obtain the given quartic polynomial as:
x^4-4n*x^2 which is factorizable as:
x^2(x^2-4n)
and, it may be observed that each of the quadratic polynomial factors has integer coefficients.
Consequently, by combining the CASES I and II we observe that:
EITHER, n= m^2, where m is any given integer.
OR, n=p
Edited on May 24, 2022, 10:57 pm
Explanation to Puzzle Answer
