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There is a prior question to show that polynomial of degree 7 with all integer coefficiants has 7 integer values $ P(x1,x2...x7)=+-1$ cant be expressed as product of two polynomials with integer coefficiants.

The solution is to try to express $P(x)=M(x)*Q(x)=+-1$ Highest degree of smaller degree polynomial lets say $M(x)$ can be 3. Given that $M(x)$ can either be $+1$ or $-1$ for $x1,x2...x7$. If $M(x)$ is of $degree 3$ and is equal $1$ in 3 cases,than it also has other 4 x's that are equal either 1 or -1.Which is impossible.It is impossible for any polynomial degree of n to have n+1 different roots.

Thats all ok,but I wonder why it was important to highlight that polynomial have integer coeffitients and 7 integer solutions which give outpus of 1 or -1. As far as I know any polynomial whather with integer or non integer coeffitiants cant have more roots than its degree.

Bill Dubuque
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Roger Powell
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  • If the coefficients weren't integers, then you might have $M(x_1)=2$ and $Q(x_1)=\frac 12$, for example. – lulu Mar 18 '24 at 23:57
  • Why is that?How does P(x) having integer coefficiants quarantees that any of its factor will have integer coefficiants – Roger Powell Mar 19 '24 at 00:04
  • Nobody said anything about constant coefficients. I am just pointing out that, if $R(x)$ is a polynomial with integer coefficients and $n$ is an integer then $R(n)$ is an integer. That's what is being used here. Well, that and the fact that $\pm 1$ have no integer factors other than $\pm 1$. – lulu Mar 19 '24 at 00:07
  • I see you edited your comment to remove the reference to "constant coefficients". But switching it to "integer coefficients" does not make sense since the problem directs us to assume that $P(x)$ and the potential factors $M(x), Q(x)$ have integer coefficients. – lulu Mar 19 '24 at 00:11
  • I get it I made small mistake in comment instead of constant I meant integer coefficiant and I edited.My question is:Polynomial P(x) have integer coefficiants why can claim that its factor polynomials N(x) and M(x) have intiger coefficiants,when polynomial with integer coefficianct can be product of polynomials with non intger coefficiants? – Roger Powell Mar 19 '24 at 00:11
  • Again, the problem directs us to consider factors with integer coefficients. – lulu Mar 19 '24 at 00:12
  • Oh you are right.My mistake.The problem assumes that M(x) and N(x) have integer coefficiants. – Roger Powell Mar 19 '24 at 00:14
  • The key idea is that if we evaluate a product of integer coef polynomials at integer values then we get a value that is a product of integers, so knowing how its integer values factor yields constraints on how the polynomial factors. In fact we can devise a polynomial factorization algorithm from such knowledge, e.g. see here in the linked dupe. – Bill Dubuque Mar 19 '24 at 00:52
  • Thats cool.It helps me to get the bigger picture.Altough I assume there is even bigger picture than that. – Roger Powell Mar 19 '24 at 14:31

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