Hmm f(x) = rx + b where r and b are rational numbers.

I did a lot of plugging in variables and eventually got y=ax+b for rational a and b

yep. on the test I accidentally allowed even degree terms

I also got $f(x) = ax+b$ for any rational a and b.

Is it just me or was this problem too easy.

EDIT: jkjk being stupid

@Octophi: It's not necessarily a polynomial, is it? You can't just assume degree. It could be strange beyond belief, like the ruler function.

Is it ok to automatically assume the function is a polynomial?

you can't assume polynomial wlog $f(0)=0$ lemma: $f(a)+f(b)=2f((a+b)/2)$ proof: $(x,y,z,w)=(a,(3a+b)/4, (a+b)/2, (a+3b)/4)$ and $((3a+b)/4, (a+b)/2, (a+3b)/4, b)$ now add plug in $a=0, b=2/q$ for integer $q$ use induction to show $f(i/q)=if(1/q)$ plug in $i=q$, we win

Yeah, I figured my solution was wrong because I couldn't assume it was a polynomial but I couldn't really think of anything else to write down so I went with that...hopefully I'll get a point or something

My solution was the following: Replace $x,y,z,t$ with $a, a+d, a+2d, a+3d$ where $d > 0$ to get \[f(a) + f(a+3d) = f(a+d) + f(a+2d).\] Now replace $a$ with $a-d,$ to get \[f(a-d) + f(a+2d) = f(a) + f(a+d).\] We can add these two equations so that terms cancel: \[f(a) + f(a+3d) + f(a-d) + f(a+2d) = f(a+d) + f(a+2d) + f(a) + f(a+d),\] which simplifies to \[f(a-d) + f(a+3d) = 2f(a+d).\] For $m, n \in \mathbb{Q},$ this is equivalent to \[f(m) + f(n) = 2f\left(\dfrac{m+n}{2}\right).\] (This is Jensen's functional equation, by the way.) I wanted to reduce this to Cauchy's functional equation somehow, which I believe is citable, but I'm still not too sure, so I set $m = 0$ and $n \to m + n$ to get $f(0) + f(m+n) = 2f((m+n)/2).$ Hence \[f(m) + f(n) = f(0) + f(m+n).\] Now just let $g(x) = f(x) - f(0),$ and substitute to get \[g(x) + g(y) = g(x+y).\] This is Cauchy's functional equation over the rationals, hence $g(x) = ax$ for some $a,$ so $f(x) = ax + b$ for some $a, b \in \mathbb{Q}.$

Benq wrote: Is it ok to automatically assume the function is a polynomial? No way! That's a huge assumption. We're looking for any functions on the rationals that satisfy this -- by assuming a polynomial you're giving yourself a ton of regularity to work with (actually, this problem is almost trivial with continuity arguments if you assume a polynomial).

Well, it was a rational function, how did you prove its a polynomial??

infiniteturtle wrote: you can't assume polynomial wlog $f(0)=0$ Wait can you do that?? Also @mjoshi: in this context, rational function doesn't mean a quotient of polynomials. It means a function whose values are always rational numbers.

yeah add a constant and it's still true iff the original was

MSTang wrote: infiniteturtle wrote: you can't assume polynomial wlog $f(0)=0$ Wait can you do that?? Just subtract $f(0)$ from the function -- let $g(x) = f(x) - f(0)$ then if you show that $g(x)$ is linear so is $f(x)$.

Does this work? oops I'm a dumby never mind wait maybe not dumby???

Hmmm, my solution was really not rigorous. How many points do you think I might get? So basically I showed that f(x)+f(y)=2f((x+y)/2), which basically shows that given 2 points on the graph of f(x), the midpoint also lies on the graph. So basically i continued by trying to construct every single rational number by constructing more and more midpoints, to show that they all lied on the same line.

My Solution: we can guess see here linear solution $f(x)=kx+c$ satifies the solution putting putting $0$ here is not reasonable because we cannot get something from that as last functional equation we seen so lets write this as something reasonable we know that this is arithmatic progression let $x=a$,$y=a+d$,$z=a+2d$ and $t=a+3d$ ofcource for $d>0$ so we can write this functional equation as \[\implies f(a)+f(a+3d)=f(a+d)+f(a+2d)\]so actually the point here is to reduce the terms (yeah i know we make lot of terms by this you will see) and make things cancel lets see what i mean umm what if we let first term of the arithmetic progression is $a-d$ lets see the equation becomes like \[\implies f(a-d)+f(a+2d)=f(a)+f(a+d)\]now you i guess can see what i want to do here so what we do now is add both of these lets see what we get. \[\implies f(a-d)+f(a+3d)=2f(a+d) \;\;\;\; \forall\,a\in\mathbb{Q} \,\, ,d>0\]let $x=a-d$ and $y=a+3d$ so its basically written as\[\implies f(x)+f(y)=2f\left(\frac{x+y}{2}\right)\]which is basically jensen's functional equation so the solution of this is $f(x)=kx+c$ and we are done$\blacksquare$

Is that equation citable? I simplified it down to Cauchy's

Evan Chen A.3 says Quote: In general it is usually okay to cite a result that is (i) named, and (ii) does not trivialize the given problem. so it should be fine.

Solved with HamstPan38825. We claim the only solutions are $f(x)=kx+c$, and it is easy to see that this works. Now, define $x=a$, $y=a+d$, $z=a+2d$, and $t=a+3d$. Putting these into the equation gives us \[f(a)+f(a+3d)=f(a+d)+f(a+2d)\]If we subtract $d$ from everything, this simplifies to \[f(a-d)+f(a+2d)=f(a)+f(a+d)\]Then, if we add the two equations together and cancel, we get \[f(a-d)+f(a+3d)=2f(a+d)\]This is Jensen's Functional Inequality, which we know has solution $f(x)=kx+c$, but only if $d\neq0$. So if $d=0$, then $a=x=y=z=t$, but this contradicts the fact that $x<y<z<t$, so $d$ can't be $0$, so we're done. $\square$

my solution was wrong

Rewritten solution: Let $P(x,y,z,t)$ denote the given assertion and $d$ be some nonzero rational number. $P(x,x+d,x+2d,x+3d): f(x)+f(x+3d)=f(x+d)+f(x+2d)$. $P(x-d,x,x+d,x+2d): f(x-d)+f(x+2d)=f(x)+f(x+d)$. Adding the equations gives $f(x-d)+f(x)+f(x+2d)+f(x+3d)=2f(x+d)+f(x)+f(x+2d)\implies f(x-d)+f(x+3d)=2f(x+d)$. Since $x$ and $d$ can be any rational numbers, this simplifies to $f(x)+f(y)=2f\left(\frac{x+y}{2}\right)$. This is Jensen's FE, but I will solve it here. Let $g(x)=f(x)-f(0)$. Then $g(x)+g(y)+2f(0)=2g\left(\frac{x+y}{2}\right)+2f(0)\implies g(x)+g(y)=2g\left(\frac{x+y}{2}\right)$ with the additional constraint that $g(0)=0$. Let $Q(x,y)$ be this new assertion. $Q(x,0): g(x)=2g\left(\frac{x}{2}\right)\implies g\left(\frac{x}{2}\right)=\frac{g(x)}{2}$. Thus, $g(x)+g(y)=g(x+y)$, so $g$ is Cauchy, which is well-known to have solutions $g(x)=kx$ for some constant $k$ over the rationals. So $f(x)=kx+f(0)$, $\boxed{f(x)=kx+c}$ for rational constants $k$ and $c$. To show that this works, we have $kx+c+kt+c=ky+c+kz+c\iff kx+kt=ky+kz$. Note if $k=0$, then constant solution obviously works. If not, then $kx+kt=ky+kz\iff x+t=y+z$, which is true since they form an arithmetic progression.

The only solutions are $\boxed{f(x) = ax + b}$ for all $x$, where $a,b\in\mathbb{Q}$. These work since \[ (ax + b) + (at + b) = (ay + b) + (az + b)\iff a(x + t - y -z) = 0,\]which is true since $x + t = y + z$. It remains to show that these solutions are the only ones. Take $\{x,y,z,t\} = \{a, a + d/2, a + d, a + 3d/2\}$ and $\{x,y,z,t\} = \{a + d/2, a + d, a + 3d/2, a + 3d\}$ separately to get the equations \[ f(a) + f(a + 3d/2) = f(a + d/2) + f(a + d)\qquad (1),\]\[ f(a + d) + f( a + 3d/2) = f(a +d/2) + f(2d)\qquad (2).\]Subtracting $(1)$ from $(2)$ gives $f(a + d) - f(a) = f(a + 2d) - f(a + d)\implies f(a + d) = \frac{f(a) + f(a + 2d)}{2}$. Using the substitution $(x,y) = (a, a+ 2d)$, we have $f\left(\frac{x + y}{2}\right) = \frac{f(x)+ f(y)}{2}$ for all $x,y\in\mathbb{Q}$. Hence, $f$ satisfies Jensen's FE over the rationals, so $f$ is linear.

The only solutions are linear $f$, or $f(x)=mx+n$ for rational $m$ and $n$, which works. Now, we show these are the only solutions. Let rational numbers $a<b<c<d<e$ be in arithmetic progression. Then \begin{align*} f(b) + f(e) &= f(c) + f(d)\\ f(b) + f(c) &= f(a) + f(d)\\ \end{align*}Subtracting, we get $f(e)-f(c) = f(c)-f(a)$, so $f$ is linear over all arithmetic progressions in $\mathbb{Q}$. In particular, $f$ is linear for all integer inputs. Let $g$ be a linear function over $\mathbb{Q}$ such that $g(x)=f(x)$ for all integral $x$. We will show that $g=f$. Let $r=\frac p q$ be a nonzero rational number, where $p$ and $q$ are nonzero integers. Since $f$ is linear over $\{ kr \mid k\in\mathbb{Z} \}$, and $f(0)=g(0)$ and $f(p) = g(p)$, we know that $f=g$ over that set, so $f(r)=g(r)$. Hence, $f=g$, as desired.

Let the average be $a$ and $d=z-a$. Then, $f(a-3d)+f(a+3d)=f(a-d)+f(a+d)$. We can plug in $a+2d$ for $a$ and we get $f(a+5d)-f(a+3d)=f(a+d)-f(a-d)$. We add these equation to get $f(a+5d)+f(a-3d)=2f(a+d)$. We can plug in $a-d$ for $a$ to get $f(a+4d)+f(a-4d)=2f(a)$. We can plug in $4d$ for $d$ to get $f(a+d)+f(a-d)=2f(a)$ for all rationals $a$ and $d$. It is easy to see by induction that $f(cx)=cf(x)-(c-1)f(0)$ for integers $c\ge 1$. This means that $f(p)=qf(\frac{p}{q})-(q-1)f(0)$ so $f(\frac{p}{q})=\frac{f(p)+(q-1)f(0)}{q}=\frac{pf(1)+(q-p)f(0)}{q}=\frac{p}{q}f(1)+(1-\frac{p}{q})f(0)=\frac{p}{q}(f(1)-f(0))+f(0)$ so $f(x)=x(f(1)-f(0))+f(0)=mx+b$ for all rational $x$. Plugging this back in, all $f(x)=mx+b$ work.

WLOG assume that $f(0)=0$. We may now consider the quantities $f(kd)$ for some integer $k$ and positive rational $d$; by induction we find that the values $f(2kd)$ actually form an arithmetic sequence. This immediately solves the problem over $\mathbb{Z}$ and subsequently $\mathbb{Q}$. $\blacksquare$ (This is essentially the same as @2above.)

Notice that for all $a$ and $d>0$, \begin{align*} f(a+d) + f(a+2d) &= f(a) + f(a+3d) \\ f(a) + f(a+d) &= f(a-d) + f(a+2d). \end{align*}Thus we have $$f(a+3d) + f(a-d) = 2f(a+d),$$so $f$ satisfies Jensen's functional equation. As a result, $f(x) = x+c$ for constants $c$ are the only solutions.

We claim only all $f(x)=mx+c, m,c\in\mathbb{Q}$ work. Note that these functions work by checking. For any $a,d$ we have $$f(a+d)+f(a+4d)=f(a+2d)+f(a+3d)$$$$f(a+2d)+f(a+5d)=f(a+3d)+f(a+4d)$$So we get $f(a+d)+f(a+5d)=2f(a+3d)$, equivalent to $$f(x-y)+f(x+y)=2f(x)$$Doing induction similar to the Cauchy Functional Equation over $\mathbb{Q}$ gives the solution set.

Same solution as many others, but I'll post here anyway.

Question: would Jensen's functional equation work here?