Simplistic Symmetric System

Okay.

From the first equation $x = f(y)$

We get $x + y = 234$

Now since : $x , y \in \mathbb{Z}^+$

Thus ,

$\begin{matrix} x & + & y & = & 234 \\ \hline 1 & + & 233 & = & 234 \\ 2 & + & 232 & = & 234 \\ \vdots & + & \vdots & = & \vdots\\ 233 & + & 1 & = & 234 \\ \end{matrix}$

$$

Thus , we'll have $\boxed{233}$ ordered pairs of $(x,y)$

$\Big[$ As count of different values of $x$ will be from $1$ to $233$ $\Big]$

Note that $f(f(x))=f(234-x)=234-(234-x)=x$ Then, for every value of $x$ , we can (and must) take $y=f(x)$ , since $x=f(f(x))$ , and $y=f(x)$ . Hence, we must simply count the positive integers $x$ such that $f(x)$ is a positive integer. This is simply the first $\boxed{233}$ positive integers.

Clearly if x=234-y and y=234-x we must have x+y=234. Since x,y are positive integers we have 233 pairs

Since $f(x)=234-x$ , we substitute; so our system becomes $x=234-y$ and $y=234-x$ . Since these two equations are equivalent, we only need to consider all possible solutions of positive integers to $x+y=234$ . Since $x$ can range from 1 to 233, and $y$ will range from 233 to 1 (respectively), then there are $\boxed{233}$ ordered pairs.

Obviosly $x = f(y) \Rightarrow x + y = 234$ which has $235$ on natural numbers (including $0$ ). But we want the POSITIVE natural numbers. So we have to subtract two solutions (when $x = 0$ or $y = 0$ ). So we have $235 - 2 = 233$ solutions.

We are given that 234-x=y and 234-y=x. This implies that x+y = 234. This, and that x and y are positive integers, are our only conditions. we are therefore looking for all the pairs of positive integers that sum to 234.

I am sure there is a better way of thinking about it, but to find how many there are I did the following:

234/2=117. This means that (117,117) works, which is unique in that (a,b)=(b,a). If it were 233, an odd number, we would not have this interesting fact.

We then simply multiply 117x2 and subtract 1 to account for (117,117), getting 233 as our answer.

The given equations to be solved are $x = 234 - y$ and $y = 234 - x$

So it is enough if we merely solve one of these. Let us take the equation $x = 234 - y$

We need x and y to be both positive integers, i.e., $0 < x , y$ and $x , y \in Z^+$ where $Z^+$ denotes the set of positive integers.

observe that $x$ is positive when $0 < y < 234$ or $y$ takes values from 1 to 233.

This means that there are 233 possible values of $y$ and hence 233 ordered pairs.

Adding the two equations, we get -

$x+y=f(x)+f(y) \ldots (eqn(i))$

But we know that $f(x) + f(y) = (234-x) + (234-y) \ldots (eqn(ii))$

Therefore from the $eqn(i)$ and $eqn(ii)$ , we get -

$x+y = (234-x) + (234-y) \Rightarrow x+y=234$

Therefore, there are $233$ ordered pairs of $(x,y)$ which satisfy $x+y=234$

here the least value we can put in the place of x is 1. Putting 1 we get 233. so the first pair is (x.y)=(1,233) similarly we can suppose that the last par would be (233,1). So in all we can have only 233 order pair of integers.

For any x in the closed intervall [0,234], the pair (x, f(x)) meets the conditions. So we have one ordered pair for every integer x in that intervall, in total we have 235 pairs. Since we look at pairs of positive integers, we must not count the pairs (234,0) and (0,234). That means we have 233 ordered pairs of positive integers that meet the condition.

From the question we can conclude that $x$ and $y$ satisfy $x+y=234$ . $x$ can be $1,2,3.......233$ and corresponding values of $y$ will be there.Total numbers from 1 to 233 = $\boxed{233}$