Finding the value of $f\left(\frac{22}{7}\right)$ if $f(x+y)+f(x-y)=2\bigl(f(x)+f(y)\bigr)$ and $f(1)=1$ The function $f(x)$ satisfying two properties as follows:
$$f(x+y)+f(x-y)=2\bigl(f(x)+f(y)\bigr)$$
$$f(1)=1$$
Then find the value of $f\left(\frac{22}{7}\right)$.
 A: Let's see.
$f(x+0)+f(x-0)=2f(x)+2f(0)$, hence $f(0)=0$.
$f(0+y)+f(0-y)=2f(0)+2f(y)=2f(y)$, hence the function is even. (Not that we'll need it much.)
$f(x+x)+f(x-x)=2f(x)+2f(x)$, hence $f(2x)=4f(x)$.
At this point we might start to suspect what function is it, but let's not run ahead.
$f(3x)=f(2x+x)+f(2x-x)-f(x)=2f(2x)+f(x)=9f(x)$.
By extending this pattern and proving it by induction, we'll express $f(nx)$ via $f(x)$ for all integer $n$.
By reversing that expression, we'll find $f(x)$ at all rational $x$.
A: Letting $x=y=0$ proves $f(0)=0$, and then letting $x=0$ proves $f(-y)=f(y)$.
Fix a $y\ne0$ and put $$x_k:=ky\>, \qquad a_k:=f(x_k)\qquad(k\in{\mathbb Z})\ .$$ We then have
$$a_{k+1}-2a_k+a_{k-1}=2f(y)\qquad(k\in{\mathbb Z})\ .\tag{1}$$
This shows that the second difference of the $a_k$ is constant. The general solution of  $(1)$ therefore is
$$a_k= k^2 f(y)+\beta k+\gamma$$
with arbitrary $\beta$ and $\gamma$. From $a_0=0$ and $a_{-k}=a_k$ it follows that necessarily $\beta=\gamma=0$, so that
$$f(ky)=a_k=k^2 f(y)\qquad(k\in{\mathbb Z})\ .$$
Since this is true for arbitrary $y\ne0$ we can proceed as follows: For arbitrary $r={k\over j}\in{\mathbb Q}$ we have
$$f(r)=f\left(k\cdot{1\over j}\right)=k^2f\left({1\over j}\right)=k^2{f(1)\over j^2}=r^2 f(1)=r^2\qquad(r\in{\mathbb Q})\ .$$
In particular $f\left({22\over7}\right)={484\over 49}$.
If we want $f$ continuous then $f(x)=x^2$ is the only solution to the problem. But there are many more solutions: Let $(e_\iota)_{\iota\in I}$ be a Hamel basis of ${\mathbb R}$, the latter considered as a vector space over ${\mathbb Q}$. We may assume that $e_0=1\in{\mathbb R}$ is one of the basis vectors. Let $f(e_0)=p_0:=1$, and choose $f(e_\iota):=p_\iota\in{\mathbb R}$ arbitrarily for all $\iota\ne0$. Then extend $f$  to all of ${\mathbb R}$ as follows: For arbitrary rational $r_k$ put 
$$f\left(\sum_{k=1}^N r_k e_{\iota_k}\right)=\sum_{k=1}^N r_k^2 \>p_{\iota_k}\ .$$
A: Note that
$$(x+y)^2+(x-y)^2=2(x^2+y^2)$$
This means that $f(x)=x^2$ satisfies the two properties, which means that $(22/7)^2$ is a possible value for $f(22/7)$.  It remains to rule out any other possibility.
The easiest way to do this is to show inductively that $f(nx)=n^2f(x)$ for all $n\in\mathbb{N}$, because we would then have
$$22^2f\left({7\over22}\right)=f\left(22\cdot{7\over22}\right)=f(7)=7^2f(1)=7^2$$
The identity $f(nx)=n^2f(x)$ is trivially true for $n=1$.  It's also true for $n=0$, since setting $x=y=0$ in the main property implies $2f(0)=f(0+0)+f(0-0)=2(f(0)+f(0))=4f(0)$, which implies $f(0)=0$.  If we now have $f(nx)=n^2f(x)$ and $f((n-1)x)=(n-1)^2f(x)$, then by induction, we have
$$\begin{align}
f((n+1)x)
&=f(nx+x)+f(nx-x)-f(nx-x)\\
&=2(f(nx)+f(x))-f((n-1)x)\\
&=2(n^2f(x)+f(x))-(n-1)^2f(x)\\
&=(2n^2+2-(n^2-2n+1))f(x)\\
&=(n^2+2n+1)f(x)\\
&=(n+1)^2f(x)
\end{align}$$
Remarks:  Note that the proof by induction requires two preceding cases, not just one.  Also, we've only shown, in effect, that $f(r)=r^2$ when $r$ is rational.  I won't swear to it, but I don't think you can say anything about $f(x)$ if $x$ is irrational, without additional hypotheses on $f$.  (Continuity would certainly be enough.)
