I provide both the official question and answer below. At the end, I present my doubts which are to be clarified.
QUESTION:
Let $ P(x)$ be a polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ with integer coefficients, such that $P(x) \cdot Q(x)$ is a polynomial in $ x^2$.
SOLUTION:
Let $ P(x)=\sum a_mx^m$
Let $ Q(x)=\sum a_{2j}x^{2j} - \sum a_{2k+1}x^{2k+1}$
Then $$ P(x) \cdot Q(x)=(\sum a_{2j}x^{2j})^2 -(\sum a_{2k+1}x^{2k+1})^2$$ is a polynomial in $ x^2$.
MY DOUBTS:
This is what the official solution claims. But the solution still remains to be a $\text {degree} > 2$ type of a Polynomial. So, what did the question actually mean? I don't get it, actually. If the solution is wrong, then kindly present a proper solution.
Perhaps it might be easier to look at things in a slightly different way.
If we start with $p(x)=e(x)+o(x)$ where $e(x)$ consists of the even powers of $x$ from $p(x)$ and $o(x)$ is likewise the odd powers, we can put $q(x)=e(x)-o(x)$.
Then $$p(x)q(x)=e(x)^2-o(x)^2$$Now every term in $e(x)$ is made up of even powers of $x$, so the same is true of $e(x)^2$ and every term in $o(x)^2$ involves the product of two odd powers of $x$ which results in an even power of $x$. So the result can be expressed as a polynomial in $y=x^2$
That is the structure of the proof you have been given, stated without looking at the individual terms. It makes it clear that what is going on involves the even and odd powers of $x$.
Now we define (or recall) an even function $e(x)$ is one for which $e(x)=e(-x)$, and an odd function has $o(x)=-o(-x)$. For an arbitrary polynomial $p(x)$ you can check that $$e(x)=\frac {p(x)+p(-x)}2; \text { and } o(x)=\frac {p(x)-p(-x)}2$$have the indicated properties. Moreover these polynomials are the ones we defined earlier by looking at the powers of $x$ explicitly (in particular the coefficients are integers, as required). And we see that $e(x)$ consists of even powers of $x$ - and hence is a function of $x^2$. And in turn a function of $x^2$ is always an even function.
But the argument here with $p(x)$ no longer depends on the fact that $p$ is a polynomial - it could be a more general function (note for the future).
Finally, since we have expressions for $e(x)$ and $o(x)$ what does $q(x)$ look like in these terms. Well $$q(x)=e(x)-o(x)=p(-x)$$
So it would have been very much easier to say this at the beginning, and to note that $p(x)p(-x)$ is clearly an even function, and as a polynomial, must be a polynomial in $x^2$.
I hope I have not made this more complicated than it need be.