Functional Equation!

It is given that $\begin{aligned} f \left(x+\frac{y}{x}\right) & = f(x) + \frac{f(y)}{f(x)} + 2y \end{aligned}$

$\begin{array} {lrll} y = x & \Rightarrow f(x+1) & = f(x) + 1 + 2x & ...(1) \\ & \Rightarrow f(x+2) & = f(x+1) + 1 + 2(x+1) \\ & & = f(x) + 4x + 4 & ...(2) \\ y = 2x & \Rightarrow f(x+2) & = f(x) + \dfrac{f(2x)}{f(x)} + 4x & ...(3) \\ (3) - (2): & 0 & = \dfrac{f(2x)}{f(x)} - 4 \\ & \Rightarrow f(2x) & = 4f(x) & ... (4) \\ (4): x = 1: & f(2) & = 4f(1) &...(4a) \\ (1): x=1: & f(2) & = f(1) + 3 & ...(1a) \\ (4a)=(1a): & 4f(1) & = f(1) + 3 \\ & f(1) & = 1 \end{array}$

From $(1)$ we have:

$\begin{aligned} f(x+1) & = f(x) + 2x + 1 \\ \Rightarrow \sum_{x=1}^n f(x+1) & = \sum_{x=1}^n (f(x) + 2x + 1) \\ & = \sum_{x=1}^n f(x) + n(n+1) + n \\ \sum_{x=1}^n f(x+1) - \sum_{x=1}^n f(x) & = n^2 + 2n \\ \sum_{x=1}^{n+1} f(x) - f(1) - \sum_{x=1}^n f(x) & = n^2 + 2n \\ \sum_{x=1}^{n+1} f(x) - \sum_{x=1}^n f(x) & = n^2 + 2n + f(1) \\ f(n+1) & = n^2 + 2n + 1 = (n+1)^2 \\ \Rightarrow f(n) & = n^2 \\ \Rightarrow f(2015) & = 2015^2 = \boxed{4060225} \end{aligned}$

We are given $\begin{aligned} f \left(x+\frac{y}{x}\right) & = f(x) + \frac{f(y)}{f(x)} + 2y \end{aligned}$

Now, plug $x=y$ , we get:

$f(x+1) = f(x) + 1 + 2x$

Now, add $x^2$ to both the sides

$f(x+1) + x^2 = f(x) + 1 + 2x + x^2$

$\Rightarrow f(x+1) + x^2 = f(x) + (x+1)^2$

$\Rightarrow f(x+1) - (x+1)^2 = f(x) - x^2$

Now, define $g(x) = f(x) - x^2$

So, we get :

$g(x+1) = g(x)$ $\Rightarrow g$ $is$ $a$ $constant$ $function$

So, let $g(x) = c$

$\Rightarrow f(x) = x^2 + c$

Plugging $f(x) = x^2 + c$ in the parent equation, we get :

$x^2 + \frac{y^2}{x^2} + 2y + c = x^2 + c + \frac{y^2 + c}{x^2 + c} + 2y$

$\Rightarrow \frac{y^2}{x^2} = \frac{y^2 + c}{x^2 + c}$

$\Rightarrow cx^2 = cy^2$

$\Rightarrow c = 0$

Hence, $f(x) = x^2$

So, $f(2015) = 2015^2 = \boxed{4060225}$

plug in x=y then get: f(x+1)=f(x)+1+2x f(x+1)-f(x)=2x+1 let n be an integer: f(n+1)-f(n)=2x+1 we know from arithmetic series that if a(n+1)-a(n)=an+b that means that a(n) can be a sum of an arithmetic series therefore is a quadradic function with no "c". one can solve the equation by defining f(n)= an^2+bn. after that one gets that f(n)=n^2 and therefore f(2015)=2015^2

Putting $y=x=n$

We have following recurrence

$f(n+1) - f(n) =2n+1$ .

Now taking summation as n varies from 1 to 2014

$f(2015)=f(1) + 2014\times 2016$ .

Put $y=2x$ in first equation and use the recurrence relation to get

$f(2x)=4 f(x)$ .

$f(2)=4f(1)$ .

Now using recurrence relation $f(2)=f(1)+3$ .

Therefore $f(1)$ = 1.

Hence the conclusion follows