MHB Expressing a polynomial P(x)=(x−a)^2(x−b)^2+1 by two other polynomials

[lfdahl]

Let $a$ and $b$ be two integer numbers, $a \ne b$. Prove, that the polynomial:

$$P(x) = (x-a)^2(x-b)^2 + 1$$

cannot be expressed as a product of two nonconstant polynomials with integer coefficients.

 

[kaliprasad]

Let $a$ and $b$ be two integer numbers, $a \ne b$. Prove, that the polynomial:

$$P(x) = (x-a)^2(x-b)^2 + 1$$

cannot be expressed as a product of two nonconstant polynomials with integer coefficients.

Spoiler

by letting t = x - a and hence x-b = t + a -b = t -c where c = b- a we get

$t^2(t-c)^2 + 1$ another polinomial in t

above has minimum value of 1 so this does not have linear factor as no real zero.

so we need to check if it has 2 quadratoc factors

1 can be expressed as 1 * 1 or - 1 * - 1 so we have 2 choices

$(t^2+nt +1)(t^2-nt + 1)$ n and -n because there is no term in t

this gives that there is no term in $t^3$ which is contradiction

similarly $(t^2+nt -1)(t^2-nt - 1)$ is ruled out

hence it cannot be factored

 

[lfdahl]

Spoiler

by letting t = x - a and hence x-b = t + a -b = t -c where c = b- a we get

$t^2(t-c)^2 + 1$ another polinomial in t

above has minimum value of 1 so this does not have linear factor as no real zero.

so we need to check if it has 2 quadratoc factors

1 can be expressed as 1 * 1 or - 1 * - 1 so we have 2 choices

$(t^2+nt +1)(t^2-nt + 1)$ n and -n because there is no term in t

this gives that there is no term in $t^3$ which is contradiction

similarly $(t^2+nt -1)(t^2-nt - 1)$ is ruled out

hence it cannot be factored

Very nice, kaliprasad! Thankyou for your participation!The suggested solution differs a bit from kaliprasads, so I will post it here:

Spoiler

Suppose $P(x)$ is a product of two polynomials. Then these polynomials are monic and both have degree $2$, since the coefficient of $P(x)$ at $x^4$ is $1$, and $P(x) > 0$ has no real roots:

\[P(x) = (x^2+px+q)(x^2+rx+s)\]

At $x=a$ and $x=b$, $P(x)=1$, therefore: $x^2+px+q=x^2+rx+s = \pm 1$. Thus $px-rx+q-s$ has two different roots, $a$ and $b$, and therefore is identically zero: $p=r$ and $q=s$.

Then, we have: $P(x) = (x-a)^2(x-b)^2 + 1=(x^2+px+q)^2$.

Since, the only two squares, that differ by $1$ are $0^2$ and $1^2$, $P(x) = 1$. A contradiction.