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Reducible polynomial + integer = Reducible polynomial ? As the title says.

More specific :

For every integer $n$, does there exist a pair of polynomials $p(x)$ and $q(x)$ such that:

  • $p(x)$ and $q(x)$ each have integer coefficients with no common factor
  • $p(x)$ and $q(x)$ are reducible polynomials
  • $p(x) + n = q(x)$.

I mean integer nonlinear polynomials without a constant multiplication factor such as $(x^2 + 5)(x+1)$ ( but not $2x^2 $which has constant factor $2$). I think an example clarifies :

$(x)(x-2) + 1 = (x-1)^2$

where $x(x-2)$ and $(x-1)^2$ are the reducible polynomials ( reducible means we can factor them over the ring of integers ) and $+1$ is the integer.

Are there other such identities for the integer $+1$ ? Are there such identities for the integer $+2$ ? ( im aware that $x(x+1) = 0$ mod 2 for integer $x$ but I did not say $x$ is an integer ) How about the integer $+3$ ? Are there such pairs of reducible polynomials for all integers $+n$ ? And if yes , how to find such ?

Edit : I used Owens answer to clarify my question. Edit2 : I ask for additional solutions. Such as higher degree polynomials. Or generalizations. Such as $p(x) + 1 = q(x)$ and $p(x) + 2 = r(x)$.

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I conjecture that Edit 2 has no solutions ? –  mick Oct 17 '12 at 21:51
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3 Answers

up vote 3 down vote accepted

Here's what I think you're asking:

For every integer $n$, does there exist a pair of polynomials $p(x)$ and $q(x)$ such that:

  • $p(x)$ and $q(x)$ each have integer coefficients with no common factor
  • $p(x)$ and $q(x)$ are reducible polynomials
  • $p(x) + n = q(x)$.

The answer is to this question is Yes. Just consider $$p(x) = x^2 - (n+1) x = (x)(x - (n + 1))$$ $$q(x) = x^2 - (n+1) x + n = (x - 1)(x - n).$$

This directly generalizes your example with $n=1$.

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Ah yes Thanks for the answer Owen. How about more cases ? –  mick Oct 11 '12 at 15:59
    
You get a very similar example for every factorization $n=k\ell$ of $n$: just take $p(x) = x^2 - (k+\ell)x = x(x-k-\ell)$ and $q(x) = x^2 - (k+\ell)x + n = (x-k)(x-\ell)$. –  Owen Biesel Oct 11 '12 at 16:03
    
Oh Nice , this is clear from the facts about sums and products of the zero's for polynomials of degree 2. –  mick Oct 11 '12 at 16:10
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We will cheat. Let $a(x)$ and $b(x)$ be any two monic polynomials of degree $\ge 1$ with the following properties. (i) $a(x)$ has constant term $1$; (ii) $b(x)$ has constant term $-n$. Let $p(x)=a(x)b(x)$ and $q(x)=a(x)b(x)+n$. Then $q(x)$ has constant term $0$, and has degree $\ge 2$, so is reducible.

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Indeed. Im surprised. Thank you. –  mick Oct 11 '12 at 20:10
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Hint $\rm\ a\!+\!a' = b\!+\!b'\:\Rightarrow\: (x\!-\!a)(x\!-\!a')-(x\!-\!b)(x\!-\!b')\,=\, aa'\!-bb'$

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@ Bill : I think you made a typo not ? –  mick Oct 11 '12 at 16:07
    
@mick Huh? e.g. your example comes from $\: 1 + 1\, =\, 0 + 2.$ –  Bill Dubuque Oct 11 '12 at 16:12
    
Your right sorry. Does this mean your answer is equivalent to Owen's ? –  mick Oct 11 '12 at 16:29
    
@Mick Owen's answer is the special case $\rm\: n + 1\, =\, 0 + (n\!\!+\!1).$ –  Bill Dubuque Oct 11 '12 at 17:06
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