How about when, for some particular value(s) of $c$, we have $c$ a divisor of $2014$? For example, for $c=2014$, we get $P(x) = \dfrac {1}{2014}x$. It is quite easy to find all these cases. Not to mention that $P(x) = 1$ does not verify the equation, for no value of $c$ (your error being that a polynomial with infinitely many roots not only is it constant, but is identically null).

Let ${x\choose n}=\frac{x(x-1)\dots (x-n+1)}{n!}$, for each $n\in \mathbb{N}, x \in \mathbb{R}$. First, one must prove that $c \ne 0$. This is easy, because if $c=0$, then $xP(x)=(x-2014)P(x)$, which impies $P(x)=0$, absurd, since $P(2014)=1$. Next, we prove that $c$ must be a divisor of $2014$, and to do this, we can use roza2010's idea. Suppose that $c$ is not a divisor of $2014$. For $x=2014$ we get $P(2014-c)=0$ For $x=2014-c$ we get $P(2014-2c)=0$ For $x=2014-2c$ we get $P(2014-3c)=0$ Inductively: For $x=2014-nc$ we get $P(2014-(n+1)c)=0$ (we can only get this because $2014-nc \ne 0, \forall n \in \mathbb{N}$). Polynomial $P$ has infinite roots, so $P(x) \equiv 0$, a contradiction, since $P(2014)=1$. With this, we get that $c$ is a divisor of $2014$. So, we can write $2014=ac$, with $a, c \in \mathbb{Z}$. For $x=2014+c=c(a+1)$ we get $P(c(a+1))=a+1$ For $x=2014+2c=c(a+2)$ we get $P(c(a+2))=\frac{(a+2)(a+1)}{2}$ Inductively, we get $P(c(a+n))=\frac{(a+n)(a+n-1) \dots (a+1)}{n!}={ a+n \choose n}={a+n \choose a}$. From here, we can see that $a<0$ is impossible, since $P(c(a+n)) = { a+n \choose n} = 0$, for $n \ge -a$. This implies that $P$ has infinitely many roots, hence $P \equiv 0$, absurd, because $P(2014)=1$. Thus, $a>0$, which means that $c$ is a positive divisor of $2014$. Now, observe that from $P(c(a+n))={a+n \choose a}$, we have that $P(x)={\frac{x}{c} \choose \frac{2014}{c}}$ for infinitely many $x$, so $P(x)={\frac{x}{c} \choose \frac{2014}{c}}$ for all $x$. This is a souction indeed, because: $(x-2014)P(x)=(x-c(\frac{2014}{c})).{\frac{x}{c} \choose \frac{2014}{c}}=$ $=c.\frac{(\frac{x}{c}-\frac{2014}{c})(\frac{x}{c})(\frac{x}{c}-1) \dots (\frac{x}{c}-\frac{2014}{c}+1)}{(\frac{2014}{c})!}=$ $=c(\frac{x}{c}).((\frac{x}{c}-1) \dots (\frac{x}{c}-\frac{2014}{c}+1)\frac{(\frac{x}{c}-\frac{2014}{c})}{(\frac{2014}{c})!}=$ $=c(\frac{x}{c}).{\frac{x}{c}-1 \choose \frac{2014}{c}}=x{\frac{x-c}{c} \choose \frac{2014}{c}}=xP(x-c)$. Finally, for $c$ being a positive divisor of $2014$, we have $P(x) \equiv {\frac{x}{c} \choose \frac{2014}{c}}$, and for any other $c$ there is no solutions.

There is an easy solution. First, we try to see if it is a constant solution: P(x) = k, since P(2014)=1 then P(x)=1, then x = 2014-x which is a contradiction. Then we re-write the original expression: P(x) = x[P(x)-P(x-c)]/2014 Since P(x) = x * (something) Now there are 2 options: 1. P(x) is constant (So P(x)-P(x-c) = 0) 2. P(x) doesn't have a constant term (or rather the constant term is 0) We will use this trick to find P(x) Suppose A(x) = P(x), then either A(x) is constant or it doesn't have a constant term. If it doesn't have a constant term then we can write it as A(x) = x *B(x) So we look back at our old expression: A(x) = x[A(x)-A(x-c)]/2014, replace A(x) with xB(x) xB(x) = x[xB(x)-xB(x-c)]/2014, we cancel the x's on both sides and we get the same expression for B(x) B(x) = x[B(x)-B(x-c)]/2014, The same as with A(x), if B(x) is not a constant then it canbe written as C(x), for which: C(x) = x[C(x)-C(x-c)]/2014, And so on untill one of our new functions is constant. Lets say K(x) is constant, so K(x) = a Then the previous function is J(x) = ax, the previous one I(x) = ax^2 (since every function is the previous one devided by x), and so on untill P(x) = ax^m, since P(x) is not constant m>0 Now we re-write the original expression as: ax(x-c)^m = a(x-2014)x^m, we divide both sides between ax, and distribute: (x-c)^m = (x^m-2014 x^(m-1)). Now we look at the constant terms of both sides of the equation (-c)^m is the constant of the first side. We will prove the other side has a nonzero constant term: c is not 0 since if it were (x-c)^m = (x^m-2014 x^(m-1) implies (x)^m = (x^m-2014 x^(m-1)), wich in turn: 2014 x^(m-1) = 0, since 2014 is not 0, we found a cointradiction. So the right side must have a nonzero constant term So (x^m-2014 x^(m-1)) has a constant term. Because m>0 x^m is not the constant term. So 2014x^(m-1) must be it. Implying m-1=0 And m=1 So P(x) = ax^1=ax Since P(2014) = 1, then 2014a=1 a=1/2014 So finally: P(x) = x/2014