Factorize $(X+1)^{101}+100$ in irreducibles in $\mathbb{Q}[X]$.

Question

As the title says I'm trying to factorize the polynomial $f(X)=(X+1)^{101}+100$ in irreducibles in $\mathbb{Q}[X]$. I got a hint stating that 101 is prime, but I don't see how that can be useful. Normally you would apply Eisenstein or the rational root theorem, but I can't seem to do that in this case. I also tried seeing what happens when you take $f(X-1)=X^{101}+100$, but again I can't use Eisenstein.

Thanks a lot!

Why can't you use Eisenstein? What's the constant term? What is the polynomial $\pmod {101}$? — lulu, Dec 23 '22 at 16:01
You can apply Eisenstein directly to the coefficients of $(X+1)^{101}+100.$ The same argument would apply to $(X+1)^p+(p-1)$ for prime $p.$ — Thomas Andrews, Dec 23 '22 at 16:01
I thought for Eisenstein you were suppose to write $(X-1)^{101}+100$ as $\sum a_iX^i$ explicitly and I can't calculate $(X-1)^{101}$ by hand. Why can I apply Eisenstein to the polynomial you just described, $(X+1)^p+(p-1)$ for $p$ prime? — Num2, Dec 23 '22 at 16:15
If $p$ is prime, all the coefficients of $(X+1)^p$ are divisible by $p$, except the leading and the constant coefficients (which of course are $1$). It is equivalent to saying that $\binom pk$ is divisible by $p$ if $1 \leq k \leq p-1$. — Zag, Dec 23 '22 at 16:19
You've reverted to $(X-1)^{101}$ in your last comment, but you clearly edited the question to $(X+1)^{101}.$ In any event, you will definitely need to know the binomial theorem. You don't need to "calculate by hand" $(X+1)^{101},$ you just need to know the form of the coefficients. — Thomas Andrews, Dec 23 '22 at 16:20
$$(p-1)+(1+x)^p=p+\binom{p}1x+\binom p2x^2+\cdots+\binom p{p-1}x^{p-1}+x^p$$ — Thomas Andrews, Dec 23 '22 at 16:27
Newton polygon of $X^{101}+100$ for $p=2$ is a line segment between $(0,2)$ and $(101,0)$ which does not cross any other integer point in between (because $\gcd(2,101)=1$), which implies the irreducibility. See also Explaining Newton Polygon for proving irreducibility of polynomial in $\mathbb{Z}[x]$ in elementary way. When is $X^n-a$ is irreducible over F? can also be used. — Sil, Dec 24 '22 at 08:40

Thomas Andrews · Answer 1 · 2022-12-23T17:02:00.980

Fundamentally, this is due to a property of binomial coefficients. If $p$ is prime and $0<k<p,$ the. $\binom pk$ is divisible by $p.$ (You might be required to prove this in a problem set, depending on what is already known.)

So from the binomial theorem, we get: $$(p-1)+(1+X)^{p}=p+\left(\sum_{k=1}^{p-1}\binom pk X^k\right)+X^p$$

And apply Eisenstein to that.

Of course, $X^N+(p-1)$ can often be shown irreducible directly from Eisenstein, for any $N,$ when $p-1$ has some prime $q\mid p-1$ and $q^2\not \mid p-1.$

For example, when $p\equiv 3\pmod 4,$ we can use $q=2.$ That covers this about half the primes.

But it is trickier when $p=101$ since there is no $q$ for $p-1.$

Factorize $(X+1)^{101}+100$ in irreducibles in $\mathbb{Q}[X]$.

1 Answers1