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PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.

This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$

This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as

if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$

How does one solve this problem this hint?

Edit: This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7

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    $\begingroup$ What if $n=2$, $a_1 = 1$ and $a_2 = -1$? $\endgroup$ – darij grinberg Feb 3 at 4:01
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Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.

But, since $\deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.

The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.

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  • $\begingroup$ Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial? $\endgroup$ – piyush_sao Feb 3 at 7:33
  • $\begingroup$ More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $P\neq Q$). $\endgroup$ – Eric Wofsey Feb 3 at 7:52

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