### FE: Brahmagupta-Fibonacci identity!!!

Posted:

**Mon Apr 03, 2017 8:51 pm**Find all functions from reals to reals so that

$$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$

holds.

$$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$

holds.

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Posted: **Mon Apr 03, 2017 8:51 pm**

Find all functions from reals to reals so that

$$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$

holds.

$$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$

holds.

Posted: **Fri May 12, 2017 1:44 am**

No one posted any solution. Sillies.

Let $P(x,y,z,t)$ denote the statement that $[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz) $

Now, $P(0,0,0,0) \Rightarrow 2f(0)(2f(0) - 1) = 0$

So we have $f(0)=\dfrac{1}{2}$ or $f(0)=0$

Suppose $f(0)=\dfrac{1}{2}$ . Then, $P(x,0,0,0) \Rightarrow f(x)=\dfrac{1}{2}$

We now only consider the case $f(0)=0$

From $P(x,0,z,0)$ and $P(0,x,z,0)$ we have $f(x)f(z)=f(xz)=f(-xz)$. Note that this implies $f(\sqrt{x}) = \sqrt{f(x)}$

Setting $z=1$ implies $f(x)=f(-x)$, and $f(1)=1$ or $f(x)=0$. If the latter is true we have another constant function. Let's consider the former case ($f(1)=1$)

Now, $P(ak,a,bk,b) \Rightarrow f(a)f(b)[f(k)+1]^2 = f(ab(k^2+1)) = f(a)f(b)f(k^2+1) \Rightarrow [f(k)+1]^2 = f(k^2+1)$

And, $P(x,1,z,1) \Rightarrow f(xz) + f(x) + f(z) + 1 = f(xz+1) + f(x-z)$. Or, $f(\sqrt{xz})^2 + f(\sqrt{x})^2 + f(\sqrt{z})^2 + 1 = f(\sqrt{xz})^2 + 2f(\sqrt{xz}) + 1 + f(x-z)$ setting $k=\sqrt{xz}$ in the previous equation. From this we obtain, $f(x-z) = (f(\sqrt{x}) - f(\sqrt{z}))^2$

Now, $P(x,y,1,1) \Rightarrow 2(f(x)+f(y)) = f(x+y) + f(x-y) \Rightarrow f(x+y) = (f(\sqrt{x}) + f(\sqrt{y}))^2$.

We define $g(x) = f(\sqrt{x})$, for positive reals $x$. Note that $g(x) \ge 0$.

Now, $g(x+y) = f(\sqrt{x+y}) = \sqrt{f(x+y)} = \sqrt{ ( f(\sqrt{x}) + f(\sqrt{y}))^2 } = g(x) + g(y)$

Thus $g$ follows Cauchy's functional equation and is bounded. So we have $g(x)=xg(1)$ for positive reals $x$. Since $g(1)=f(1)=1$, this implies that $g(x)=x$.

Thus $f(x) = f(\sqrt{x})^2 = g(x)^2 = x^2$ for all $x$ (Since $f(-x)=f(x)$ and $f(0)=0$).

So the only functions functions which could fit the given equation are $f(x)=\dfrac{1}{2}$ or $f(x)=0$ and $f(x)=x^2$. Verifying the first two is trivial, and verifying the third leads, indeed, to the Brahmagupta-Fibonacci Identity which holds. So the three are the required solutions.

Let $P(x,y,z,t)$ denote the statement that $[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz) $

Now, $P(0,0,0,0) \Rightarrow 2f(0)(2f(0) - 1) = 0$

So we have $f(0)=\dfrac{1}{2}$ or $f(0)=0$

Suppose $f(0)=\dfrac{1}{2}$ . Then, $P(x,0,0,0) \Rightarrow f(x)=\dfrac{1}{2}$

We now only consider the case $f(0)=0$

From $P(x,0,z,0)$ and $P(0,x,z,0)$ we have $f(x)f(z)=f(xz)=f(-xz)$. Note that this implies $f(\sqrt{x}) = \sqrt{f(x)}$

Setting $z=1$ implies $f(x)=f(-x)$, and $f(1)=1$ or $f(x)=0$. If the latter is true we have another constant function. Let's consider the former case ($f(1)=1$)

Now, $P(ak,a,bk,b) \Rightarrow f(a)f(b)[f(k)+1]^2 = f(ab(k^2+1)) = f(a)f(b)f(k^2+1) \Rightarrow [f(k)+1]^2 = f(k^2+1)$

And, $P(x,1,z,1) \Rightarrow f(xz) + f(x) + f(z) + 1 = f(xz+1) + f(x-z)$. Or, $f(\sqrt{xz})^2 + f(\sqrt{x})^2 + f(\sqrt{z})^2 + 1 = f(\sqrt{xz})^2 + 2f(\sqrt{xz}) + 1 + f(x-z)$ setting $k=\sqrt{xz}$ in the previous equation. From this we obtain, $f(x-z) = (f(\sqrt{x}) - f(\sqrt{z}))^2$

Now, $P(x,y,1,1) \Rightarrow 2(f(x)+f(y)) = f(x+y) + f(x-y) \Rightarrow f(x+y) = (f(\sqrt{x}) + f(\sqrt{y}))^2$.

We define $g(x) = f(\sqrt{x})$, for positive reals $x$. Note that $g(x) \ge 0$.

Now, $g(x+y) = f(\sqrt{x+y}) = \sqrt{f(x+y)} = \sqrt{ ( f(\sqrt{x}) + f(\sqrt{y}))^2 } = g(x) + g(y)$

Thus $g$ follows Cauchy's functional equation and is bounded. So we have $g(x)=xg(1)$ for positive reals $x$. Since $g(1)=f(1)=1$, this implies that $g(x)=x$.

Thus $f(x) = f(\sqrt{x})^2 = g(x)^2 = x^2$ for all $x$ (Since $f(-x)=f(x)$ and $f(0)=0$).

So the only functions functions which could fit the given equation are $f(x)=\dfrac{1}{2}$ or $f(x)=0$ and $f(x)=x^2$. Verifying the first two is trivial, and verifying the third leads, indeed, to the Brahmagupta-Fibonacci Identity which holds. So the three are the required solutions.