Functions are awesome!!! -Part 3

Let $P(x,y)$ be the equation $f(x+y) f(x-y) = (f(x) + f(y))^2 - 4x^2 f(y)$ .

$P(0,0)$ gives $f(0)^2 = (2 f(0))^2 = 4 f(0)^2$ and so $f(0)^2 = 0 \implies f(0) = 0$ .

Now $P(x,x)$ gives $0 = f(2x)(0) = (2f(x))^2 - 4x^2 f(x) = 4f(x)^2 - 4x^2 f(x) = 4 f(x) (f(x) - x^2)$

So for each $x$ , either $f(x) = 0$ or $f(x) = x^2$ .

Case 1: Suppose that there is some $a \neq 0$ such that $f(a) = 0$ .

Then $P(x,a)$ gives $f(x+a)f(x-a) = f(x)^2$ for all $x$ .

If there is any $b$ such that $f(b) \neq 0$ , then $f(b) = b^2 \neq 0$ .

$b^4 = f(b)^2 = (b+a)(b-a) = (b+a)^2 (b-a)^2 = (b^2 - a^2)^2 = b^4 - 2b^2a^2 + a^4$

Thus, $2b^2a^2 = a^4$ and since $a \neq 0$ , we have $b = \pm \frac{a}{\sqrt{2}}$ .

So except for possibly these two points, $f(x) = 0$ everywhere. Now using any point other than $a$ , the same argument will also rule out those two points. So $f(x) = 0$ everywhere.

Case 2: $f(x) \neq 0$ whenever $x \neq 0$ . Then for all nonzero $x$ , $f(x) = x^2$ . So $f(x) = x^2$ everywhere.

Checking solutions: If $f(x) = 0$ , then $P(x,y)$ becomes: $(0)(0) = (0 + 0)^2 - 4(0)^2 (0)$ which is true.

If $f(x) = x^2$ , then $P(x,y)$ becomes: $(x+y)^2 (x-y)^2 = (x^2 + y^2)^2 - 4x^2 y^2,$ which is true because:

$(x^2 + y^2)^2 - 4x^2 y^2 = x^4 + 2x^2y^2 + y^4 - 4x^2y^2 = x^4 - 2x^2y^2 + y^4 = [x^2 - y^2]^2 = [(x+y)(x-y)]^2 = (x+y)^2 (x-y)^2 .$

Conclusion: There are exactly two functions that satisfy $P(x,y)$ : $f(x) = 0$ and $f(x) = x^2$ .

My solution is somewhat similar, except for the last bit:

1) Consider $(0,0)$ . It follows that $f(0)^2=4f(0)^2⟹ f(0)=0$

2) Consider $(x,x):$

$f(2x)f(0)=4⋅f(x)⋅(f(x)-x^2 )$

From $f(0)=0$ it follows $f(x)⋅(f(x)-x^2 )=0$

Thus, we must have: $f(x)=0∨ f(x)=x^2$ for all $x$ .

3) Suppose that $f(a)=0$ for some $a≠ 0$ . Consider $(x,a)$ , and $(a,x)$ Plugging in yields:

$f(x+a)^ ⋅f(x-a)=f(x)^2$

$f(a+x)^ ⋅f(a-x)=f(x)^2-4x^2 f(x)$

Notice that $f(x-a)=f(a-x)$ therefore subtracting the 2 equations: $0 = 4x^2⋅f(x)$ meaning $f(x)=0$ everywhere.

4) Thus, the only solutions are $f(x)=0$ and $f(x)=x^2$