An algebra problem by Umang Gusain

correct me if I'm wrong but I suggested that we take the given function and swap ${x}$ with ${x}$ + ${y}$ and ${y}$ with ${x}$ - ${y}$ to get ${f}$ ( ${x}$ , ${y}$ ) = $x^{2}$ - $y^{2}$ and ${f}$ ( ${y}$ , ${x}$ ) = $y^{2}$ - $x^{2}$ , to get an arithmetic mean of $\frac{f(x,y) +f(y,x)}{2}$ = $\boxed{0}$

Let, $x+y=n$ .....1 \ (x-y=m) .....2

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From 1 and 2, $x=\frac{n+m}{2}$ $y=\frac{n-m}{2}$

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By putting values in terms of n and m in function, $f(n,m)=\frac{n^{2}-m^{2}}{4}$

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Now,
$f(x,y)=\frac{x^{2}-y^{2}}{4}$ $f(y,x)=\frac{y^{2}-x^{2}}{4}$

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So, $A.M.=\frac{f(x,y)+f(y,x)}{2}=0$

The arithmetic mean is independent of the $(x, y)$ pair we choose so we can choose to find the mean of $f(0,0) = \boxed{0}$ , where we use $x = 0, y = 0$ to solve for $x+y=0$ and $x-y=0$ .

Another way is to find such a function $f$ like the following:

$f(x, y) = \frac{(x+y)(x-y)}{4}$

and then note that swapping $x$ and $y$ yields the negative, that is:

$f(x, y) = -f(y, x)$

and therefore the arithmetic mean is $\boxed{0}$ .

In fact you can use this intuition without finding a function $f$ by realizing that swapping the inputs is equivalent to negating $y$ which negates the entire function. Hence the mean must be 0.