Let $f: \mathbb Z \to \mathbb Z $ be a function such that $$f(34f(x)+78)=57$$

  1. What is $f(2016)$?

  2. Do we know any other value of $f$ for sure?

This is one of my own problems.


3 Answers 3


So, first of all

take any $x$ and write $y=34f(x)+78$; then $f(y)=57$ and $f(34f(y)+78)=57$; since $f(y)=57$ the latter says $f(2016)=57$.

Does f have to map everything to 57? Well,

$$f(x) = \begin{cases} 57 && \mathrm{if} \, x \equiv 78 \mod 34 \\ 0 && \mathrm{otherwise} \end{cases}$$
Then it is clearly true that $f(34f(x)+78)=57$ for any $x$. So no, $f$ needn't be constant.


that doesn't tell us whether $f(2016)$ is the only value that's required to be $57$. There certainly must be other values of $f$ that equal $57$. To see this, suppose $f(x)$ isn't $57$ (if we can't find an $x$ with this property then $f$ is always 57); then $f(34f(x)+78)=57$ but $34f(x)+78$ isn't $2016$.


we can construct solutions as follows. Pick any $t$ that isn't $57$. Now set $f(2016)=f(34t+78)=57$ and $f(\text{everything else})=t$. This satisfies the given condition because $f(x)$ is always either $57$ or $t$, and hence $34f(x)+78$ is always either $2016$ or $34t+78$, and hence $f(34f(x)+78)=57$ as required. But any two of these functions have different values of $34t+78$ and hence disagree everywhere except at $f(2016)$. Therefore, we do not know the value of any other $f(x)$.

  • $\begingroup$ i think f( f(x) ) is a recursion. in this example, i see x= 34f(x)+78 $\endgroup$
    – lois6b
    Commented Aug 9, 2016 at 11:36
  • $\begingroup$ I'm sorry, but I'm not sure what you mean. Could you be more specific about what you think I've said that's wrong? I don't think I've said x=34f(x)+78 anywhere... $\endgroup$
    – Gareth McCaughan
    Commented Aug 9, 2016 at 11:37
  • 1
    $\begingroup$ A recursive definition is one that tells you f(x) in terms of other f(...) things. This isn't really a property of the function but of how you write down its definition. E.g., consider the Fibonacci numbers: f(0)=0, f(1)=1, f(n)=f(n-1)+f(n-2). That's a recursive definition but you can define the same function by $f(n)=\frac{1}{2\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)$. $\endgroup$
    – Gareth McCaughan
    Commented Aug 9, 2016 at 11:49
  • 1
    $\begingroup$ So, anyway, what we have in the problem statement here isn't a definition at all. Rather, we're being asked what we can say about a function given a fact about it: that for any x, f(34f(x)+78)=57. It turns out that that's nowhere near enough to define the function; there are lots of functions satisfying that condition. $\endgroup$
    – Gareth McCaughan
    Commented Aug 9, 2016 at 11:52
  • 4
    $\begingroup$ So it doesn't really make sense to say that it is or isn't recursive; being recursive is a property of a definition of a function, and we don't have one. (Or, relatedly, being recursive is a property of an implementation of a function on a computer. We don't have one of those here either.) $\endgroup$
    – Gareth McCaughan
    Commented Aug 9, 2016 at 11:52

f(n) = 57
1. f(2016) = 57
2. Yes, we know all the values, they are 57. :p

  • $\begingroup$ one doubt if f(x) is number f(34*f(x)+78) will be f(some number) which is equal to 57 , how it is a constant function ? $\endgroup$
    – Amruth A
    Commented Aug 9, 2016 at 11:14
  • 3
    $\begingroup$ because the result is not x dependant. no matter the content of "f", the answer is 57 $\endgroup$
    – lois6b
    Commented Aug 9, 2016 at 11:16
  • 2
    $\begingroup$ Are you suggesting that this is the only solution? Any explanations supporting this idea? $\endgroup$
    – elias
    Commented Aug 9, 2016 at 11:17
  • $\begingroup$ i saw "f( g(x) )= whatever". but never (i think) "f( f(x) ) = whatever" is like recursion $\endgroup$
    – lois6b
    Commented Aug 9, 2016 at 11:22
  • 3
    $\begingroup$ No specific reason, it was just an example of another function that works. $\endgroup$
    – Meiffert
    Commented Aug 9, 2016 at 11:35

Partial answer




$f$ has $57$ in its range, because of the definition. (But we still have to show a number, for which it picks up this value.)


$f(2016)=f(34\times f(y)+78)$ for some $y$, thus it equals 57

  • $\begingroup$ Just let $y=34f(x)+78$ for any $x$, then $f(y)=57$ by definition. $\endgroup$
    – f''
    Commented Aug 9, 2016 at 11:25

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