For polynomials with integer coefficients $f(x)$, when is $\sqrt{f(x)+f'(x)}$ also a polynomial with integer coefficients? Let $F$ be the family of polynomial functions with integer coefficients. When is it true that $f(x)\in F$ and $\sqrt{f(x)+f'(x)}\in F$?
There are, of course, functions which meet this criteria. The most basic example is $f(x)=x^2+1$, since $f'(x)=2x$ and thus $\sqrt{f(x)+f'(x)}=\sqrt{x^2+1+2x}=\sqrt{x^2+2x+1}=\pm (x+1)$. In fact, for any $k\in\mathbb{Z}$, $k^2(x^2+1)$ meets this criteria. Are there any other functions with this property?
 A: The polynomial equation $f+f'=g$ can be solved for $f$ by repeatedly differentiating then telescoping, and the solution is $f = g - g' + g''-\dots + (-1)^n g^{(n)}$ where $n = \deg g$.
Taking $g$ to be a perfect square results in an $f$ such that $\sqrt{f+f'}$ is (the absolute value of) a polynomial.
For example, $g(x)=(x+a)^2=x^2+2ax+a^2$ gives $f(x)=(x^2+2ax+a^2)-(2x+2a)+2$ $=x^2+2(a-1)x+a^2-2a+2=(x+a-1)^2+1$.
A: To answer the question at the end of the post:  yes, there are an infinite number.  Given any polynomial $g(x)$ such that $\sqrt{f(x) + f'(x)} = g(x)$, you can view this defining equation as an ODE.  This is a linear first-order ODE for which we can write down the general solution:
$$f = e^{-x} \int g^2(x) e^{x} dx + A e^{-x}, 
$$
where $A$ is a constant defined by the initial conditions. The solution with $A = 0$ will be a polynomial solution for $f$.
So, for example, if we want to find an $f(x)$ such that $\sqrt{f + f'} = x^2 - x - 1$, then applying the above procedure yields
$$
f = e^{-x} \int (x^2 - x - 1)^2 e^x \, dx = e^{-x} \left[ (x^4 - 6x^3 + 17x^2 -32 x + 33) e^{x} \right] \\= x^4 - 6x^3 + 17x^2 -32 x + 33
$$
(thank you, Wolfram Alpha.)
A: Let $f(x) = \sum_{j=0}^{n}a_{j}x^j$.
Note $f(x) + f'(x)$ is
$$a_{n}x^n + \sum_{j=0}^{n-1}(a_{j}+(j+1)a_{j+1})x^j$$
If the above polynomial is equal to some other polynomial $p$ with coefficients in $\mathbb{Z}$ then $deg(p) = n$ and if
$$p(x) =\sum_{j=0}^{n}b_{j}x^{j}$$
we have
$$a_{n} = b_{n} \in \mathbb{Z}$$ and
$$a_{n-k-1} = b_{n-k-1} - (n-k)a_{n-k} \in \mathbb{Z} \text{ by induction}$$
Thus for every polynomial $p$ there exists a unique polynomial with integer coefficients $f_{p}$ so that $f_{p}+f'_{p} = p$.
Hence we have that for every polynomial $q$ with $q \geq 0$ there exists a unique $f_{q^2} \in F$ so that $f_{q^2}+f'_{q^2} = q^2$ so that we have
$$\sqrt{f_{q^2}+f'_{q^2}} = |q| = q$$
The pairs $(q, f_{q^2})$ form all the solutions ($q\geq 0$).
A: There are many polynomials $f(x) \in F$ for which the desired condition holds.
Consider it thusly: given $g(x) \in F$,
can you find $f(x) \in F$ such that $f(x)+f'(x)=g^2(x)$?
With $n$ being the order of $g^2(x)$,
let $G_i$ be the coefficients of $g^2(x)$
and let $f_i$ be the coefficients of $f(x)$.
Then:
$$g^2(x)=\sum_{i=0}^n G_i x^i$$
$$f(x)=\sum_{i=0}^n f_i x^i$$
$$f'(x)=\sum_{i=0}^n if_i x^{i-1}$$
Gathering like powers of $x$,
$f(x)+f'(x)=g^2(x)$ gives us:
$$f_i + (i+1)f_{i+1} = G_i \;\;\forall_i $$
Thus starting with $f_{n+1}=0$, we can find that for all $0 \le i \le n$
$$f_i = G_i - (i+1)f_{i+1}$$
Thus for any $g\in F$ there is an $f \in F$ such that $f(x)+f'(x)=g^2(x)$.
Hence there are an infinitude of polynomials satisfying your original conditions.
Given a polynomial $f(x) \in F$, however, I know of no simple way of determining whether or not there exists a corresponding $g(x) \in F$
other than computing $f(x)+f'(x)$ and see if can be factored into a square.
