A pair of functions $f(x)$ and $g(x)$ is called a Payneful pair if

(i) $f(x)$ is a real number for all real numbers $x$,

(ii) $g(x)$ is a real number for all real numbers $x$,

(iii) $f(x+y)=f(x) g(y)+g(x) f(y)$ for all real numbers $x$ and $y$,

(iv) $g(x+y)=g(x) g(y)-f(x) f(y)$ for all real numbers $x$ and $y$, and

(v) $f(a) \neq 0$ for some real number $a$.

For every Payneful pair of functions $f(x)$ and $g(x)$ :

(a) Determine the values of $f(0)$ and $g(0)$.

(b) If $h(x)=(f(x))^{2}+(g(x))^{2}$ for all real numbers $x$, determine the value of $h(5) h(-5)$.

(c) If $-10 \leq f(x) \leq 10$ and $-10 \leq g(x) \leq 10$ for all real numbers $x$, determine the value of $h(2021)$.
(a) Using $f(x+y)=f(x) g(y)+g(x) f(y)$ with $x=y=0$, we obtain $f(0)=2 f(0) g(0)$.

From this, we get $f(0)(2 g(0)-1)=0$.

Thus, $f(0)=0$ or $g(0)=\frac{1}{2}$.

Using $g(x+y)=g(x) g(y)-f(x) f(y)$ with $x=y=0$, we obtain $g(0)=(g(0))^{2}-(f(0))^{2}$.

If $g(0)=\frac{1}{2}$, we obtain $\frac{1}{2}=\frac{1}{4}-(f(0))^{2}$ which gives $(f(0))^{2}=-\frac{1}{4}$.

Since $f(0)$ is real, then $(f(0))^{2} \geq 0$ and so $(f(0))^{2} \neq-\frac{1}{4}$.

Thus, $g(0) \neq \frac{1}{2}$, which means that $f(0)=0$.

Since $f(a) \neq 0$ for some real number $a$, setting $x=a$ and $y=0$ gives

$$
f(a+0)=f(a) g(0)+g(a) f(0)
$$

Since $f(0)=0$, we obtain $f(a)=f(a) g(0)$.

Since $f(a) \neq 0$, we can divide by $f(a)$ to obtain $g(0)=1$.

Therefore, $f(0)=0$ and $g(0)=1$.

(We note that the functions $f(x)=\sin x$ and $g(x)=\cos x$ satisfy the given conditions, so there does exist at least one Payneful pair of functions.)
(b) Solution 1

Let $t$ be a real number.

From (a), $f(0)=0$ and $g(0)=1$. Thus, $(f(0))^{2}+(g(0))^{2}=1$ and so

$$
\begin{aligned}
1= & (f(t+(-t)))^{2}+(g(t+(-t)))^{2} \\
= & (f(t) g(-t)+g(t) f(-t))^{2}+(g(t) g(-t)-f(t) f(-t))^{2} \\
= & (f(t))^{2}(g(-t))^{2}+2 f(t) g(-t) g(t) f(-t)+(g(t))^{2}(f(-t))^{2} \\
& \quad+(g(t))^{2}(g(-t))^{2}-2 g(t) g(-t) f(t) f(-t)+(f(t))^{2}(f(-t))^{2} \\
= & (f(t))^{2}(g(-t))^{2}+2 f(t) f(-t) g(t) g(-t)+(g(t))^{2}(f(-t))^{2} \\
& \quad+(g(t))^{2}(g(-t))^{2}-2 f(t) f(-t) g(t) g(-t)+(f(t))^{2}(f(-t))^{2} \\
= & (f(t))^{2}(g(-t))^{2}+(g(t))^{2}(g(-t))^{2}+(f(t))^{2}(f(-t))^{2}+(g(t))^{2}(f(-t))^{2} \\
= & (g(-t))^{2}\left((f(t))^{2}+(g(t))^{2}\right)+(f(-t))^{2}\left((f(t))^{2}+(g(t))^{2}\right) \\
= & \left((f(t))^{2}+(g(t))^{2}\right)\left((f(-t))^{2}+(g(-t))^{2}\right) \\
= & h(t) h(-t)
\end{aligned}
$$

Since $h(t) h(-t)=1$ for every real number $t$, setting $t=5$ gives $h(5) h(-5)=1$, as required.

## Solution 2

Let $t$ be a real number. Then

$$
\begin{aligned}
h(t) h(-t) & =\left((f(t))^{2}+(g(t))^{2}\right)\left((f(-t))^{2}+(g(-t))^{2}\right) \\
& =(f(t))^{2}(f(-t))^{2}+(f(t))^{2}(g(-t))^{2}+(g(t))^{2}(f(-t))^{2}+(g(t))^{2}(g(-t))^{2}
\end{aligned}
$$

Since $f(0)=0$, then $f(t+(-t))=0$ which gives

$$
0=f(t) g(-t)+g(t) f(-t)
$$

and so $f(t) g(-t)=-g(t) f(-t)$.

Since $g(0)=1$, then $g(t+(-t))=1$ which gives $g(t) g(-t)-f(t) f(-t)=1$.

Squaring both sides, we obtain

$$
\begin{aligned}
1= & (g(t) g(-t)-f(t) f(-t))^{2} \\
= & (g(t))^{2}(g(-t))^{2}-2 g(t) g(-t) f(t) f(-t)+(f(t))^{2}(f(-t))^{2} \\
= & (f(t))^{2}(f(-t))^{2}-f(t) g(-t) g(t) f(-t)-f(t) g(-t) g(t) f(-t)+(g(t))^{2}(g(-t))^{2} \\
= & (f(t))^{2}(f(-t))^{2}-f(t) g(-t)(-f(t) g(-t))-(-g(t) f(-t)) g(t) f(-t)+(g(t))^{2}(g(-t))^{2} \\
& \quad \quad \quad \quad \text { using } f(t) g(-t)=-g(t) f(-t) \text { twice) } \\
= & (f(t))^{2}(f(-t))^{2}+(f(t))^{2}(g(-t))^{2}+(g(t))^{2}(f(-t))^{2}+(g(t))^{2}(g(-t))^{2} \\
= & h(t) h(-t)
\end{aligned}
$$

from above. Note that this is true for every real number $t$.

Setting $a=5$, we obtain $h(5) h(-5)=1$, as required.
(c) From (b), setting $t=2021$, we obtain $h(2021) h(-2021)=1$.

We will show that $h(2021)=1$.

Since $h(2021) h(-2021)=1$, then $h(2021) \neq 0$ and $h(-2021) \neq 0$.

Suppose that $h(2021) \neq 1$.

Since $h(x)=(f(x))^{2}+(g(x))^{2}$, then $h(x) \geq 0$ for all real numbers $x$.

Also, since $-10 \leq f(x) \leq 10$ and $-10 \leq g(x) \leq 10$ for all real numbers $x$, then

$$
h(x)=(f(x))^{2}+(g(x))^{2} \leq 10^{2}+10^{2}=200
$$

for all real numbers $x$.

Since $h(2021) \neq 1$ and $h(2021) \neq 0$, then either $h(2021)>1$ and $01$ and $01$ or $h(-2021)>1$.

If $h(2021)>1$, let $b=2021$ and $c=h(2021)$.

If $h(-2021)>1$, let $b=-2021$ and $c=h(-2021)$.

Since $c>1$, then $c^{m}$ grows without bound as $m$ increases.

In particular, $c^{m}>200$ when $m>\log _{c} 200$.

From above, $h\left(2^{n} b\right)=c^{2^{n}}$ for every positive integer $n$.

Therefore, when $2^{n}>\log _{c} 200$, we have $h\left(2^{n} b\right)>200$, which is not possible since we saw above that $h(x) \leq 200$ for all real numbers $x$.

This is a contradiction, which means that the assumption that $h(2021) \neq 1$ cannot be true, and so $h(2021)=1$. (In fact, modifying the argument above shows that $h(x)=1$ for all real numbers $x$.)