American Mathematical Monthly, October 1962:
"Let $P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $Q(x)$ with real coefficients such that $P(x)Q(x)$ has terms that are all of a degree divisible by $10^9$."
Wondering if my solution/reasoning here is correct, as it feels a bit simplistic and doesn't really use the real coefficients condition.
First let's replace $10^9$ with arbitrary $k$. $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$ is the given polynomial, and $Q(x)=b_mx^m+b_{m-1}x^{m-1}+...+b_0$ is the polynomial we are free to choose. We want $C(x)=P(x)Q(x)=c_{m+n}x^{m+n}+...+c_0$ where $c_i = 0$ if $k\nmid i$.
We get the set of relations
$c_0=a_0b_0$
$c_1=a_0b_1+a_1b_0$
...
$c_{m+n}=a_nb_m.$
But since we can choose $m$ to be arbitrarily large, we can guarantee there are more degrees of freedom ($b_i$'s) than there are constraints on the $b_i$'s. That is, there are $k-1$ constraints of the form $c_i=0$ for every $k$ coefficients $c_i$. If we increase $m$, the degree of our free polynomial, by some natural number $d$, we have an extra $d$ degrees of freedom in the form of extra $b_i$'s to play with. Meanwhile the number of extra $c_i$ coefficients in the product is also $d$, but the number of extra constraints would be something roughly like $d-\frac{d}{k}$. So we eventually have enough freedom to choose the $b_i$'s such that the constraints are satisfied. Also, the constraint equations only involve addition/subtraction/multiplication/division so we can keep the numbers real.
This seems a bit too straightforward. I've written this out in matrix form, and feel like there might be some way this could fail.
A minor remark to the reference you give. It would be helpful if you add a page number and or article/problem number, because it is rather difficult to find this in a journal of more than 100 pages per issue.
In principle you are correct and this approach will work. You would however need to demonstrate that the process indeed results in a proper solution for certain values of $m$. Moreover, for a proper proof you need to be a bit more precise in formulation. "roughly like" is something to be used in more of a sketch of an approach rather than a proof.
The first observation is that not all values of $m$ are possible. In fact the question implies that the polynomial $C(x)$ is of a degree that is a multiple of $k$. If you would like to improve your approach, I would recommend to work out a few simple examples like $P(x)=x+1$ or $P(x)=(x+1)(x+2)$ and small values of $k=2,3$. You will find when and how the coefficients $b_i$ are determined and also that those coefficients $c_i$ that are non-zero cannot be chosen freely.
Below I will sketch a different approach, which also allows you to explicitly construct the polynomials $Q(x)$ and $C(x)$, and in particular that there is always a solution such that the degree of $C(x)$ is $n k$.
Suppose we consider $P(x)=x-a$ for some constant $a$, then a possible solution is given by $Q(x)=\sum_{i=0}^{k-1} x^i a^{k-1-i}$, because $C(x) = P(x) Q(x) = x^k - a^k$.
More in general for an arbitrary polynomial of degree $n$: $P(x)=\prod_i^n (x - a_i)$ there is a polynomial $Q(x)$ of degree $n(k-1)$ such that $C(x)=P(x) Q(x) = \prod (x^k - a_i^k)$ which only contains powers of $x$ that are multiples of $k$.
If all roots of $P(x)$ were real we would be done, but in general that will not be the case. However, with $P(x)$ having real coefficients we know that if the complex value $a$ is a root then also its complex conjugate $a^*$ is a root. I leave it as an exercise to show that also with complex pairs of roots the polynomials $Q(x)$ and $C(x)$ will have only real valued coefficients.
As a final remark: the degree of $C(x)$ need not be more than $n k$ but it can be smaller. As a simple example consider $P(x)=(x-1)(x+1)$ and $k=4$. Then we would get from the aforementioned method $C(x) = (x^4 - 1)(x^4 -1)$ of degree 8, but we also know that $C(x) = (x^4 - 1)$ is a correct solution.