The Functional Equation that Trolled the NIMO Participants

First of all, the degree of the LHS is clearly the same as $\deg(f(x))$ , which I will call $d$ . The degree of the RHS is $2d - 1$ , so $d = 2d - 1 \implies d = 1$ . Now that we know that $f(x)$ is linear, we can replace $f(x)$ with $ax + b$ :

$(a(x - 1) + b) + (ax + b) + (a(x + 1) + b) = 3(ax + b) = \frac{(ax + b)^2}{2013x}$

Now we have to evaluate $ax + b$ at $x = 1$ :

$3(a + b) = \frac{(a + b)^2}{2013} \implies 3 = \frac{a + b}{2013} \implies a + b = f(1) = 6039$

(Note I divided by $(a + b)$ , so $a + b = 0$ must be taken into account. If this is the case however, it does not affect the sum of all possible values of $f(1)$ .

So the remainder of 6039 when divided by 1000 is simply the final 3 digits, $\fbox{39}$ .

Let the degree of $f(x)$ be $a$ . Then $a=\frac{a^{2}}{1}$ implies that $a$ is $1$ , i.e., let $f(x)=mx+b$ for some $m$ and $b$ . Then, the left-hand side of the equation becomes $m(x-1)+b+mx+b+m(x+1)+b=3(mx+b)=3f(x)$ . Now we substitute $1$ : $3f(1)=\frac{{f(1)}^{2}}{2013}$ , which one can simplify and factor into $f(1)(f(1)-6039)=0$ , so $f(1)=0$ or $6039$ . Therefore $N=0+6039=6039$ , and the remainder when $N$ is divided by $1000$ is $\boxed{039}$ .

Let $d = \text {deg} (f)$ . Then, the degree of the LHS is $d$ and the degree of the RHS is $2d - 1$ , so $d = 1$ . Thus, $f(x) = mx + b$ for some $m, b$ . In fact, note that $b = 0$ , so $f(x) = mx \implies 3mx = \dfrac {m^2x^2}{2013x} \implies m = 6039,$ so the answer is $\boxed {39}$ .

We have for x != 0 (! stands for not)

2013 x (f(x-1) + f(x) + f(x+1)) = f(x)^2 .............. (i)

Let d be the degree of f(x).

Using (i) we note that

1 + d = 2 d which implies that d = 1

Let f(x) = a x + b. Then, using (i) we have that

(ax + b)^2 = 2013 x (a(x-1) + b + ax + b + a(x+1) + b) = 6039 x (ax + b) ............(ii)

Since f is a non constant polynomial, ax + b is not identically equal to 0.

So, we can cancel it from (ii) , thus getting us

ax + b = 6039x i.e. a = 6039 and b = 0.

Now, a = 6039 leaves a remainder of 39 when divided by 1000.

Hence, 39 is the required answer.

To find the degree of this function, we multiply both sides by $2013x$ , giving $2013x(f(x-1)+f(x)+f(x+1))=f(x)^2$ . Assuming that the degree of $f(x)$ is $n$ , the degree of the RHS is $2n$ and the LHS is $n+1$ . Setting these to be equal, we get $n=1$ , and so $f(x)$ is linear and in the form of $mx+b$ . Plugging this into our original equation, we get (after some simplifying) 3mx+3b=\frac{f(x)^2}{2013x}, and since we know \(f(x)=mx+b , we have $3f(x)=\frac{f(x)^2}{2013x}$ . Dividing by $f(x)$ and multiplying by $2013x$ gives $f(x)=6039x$ , $f(1)=6039$ and so our answer is $39$ .

function square when divided by x gives same degree => no constant term & 1st degree => f(x) = kx

Indeed, this did troll me. "polynomials" not "functions," thanks a lot, WOOT

Note that for all $f(x)$ such that $\text{deg }(f(x))\ge 2$ , $\text{deg }(LHS)<\text{deg }(RHS)$ .

Therefore either $\text{deg }(f(x))=0,1$ . Since we know that $f(x)$ is not constant, $\text{deg }(f(x))=1$ .

We let $f(x)=ax+b$ . Plugging in to the equation, we see that $3ax+3b=\dfrac{a^2x^2+2abx+b^2}{2013x}$ , or $6039ax+6039b=a^2x^2+2abx+b^2$ .

Setting coefficients equal, we get that $6039a=a^2$ or $a=6039$ . Also, $b^2=0$ so $b=0$ . Therefore our only polynomial that works is $f(x)=6039x$ .

Plugging in $x=1$ and taking $\mod{1000}$ , we get the answer of $\boxed{39}$ .