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Let $f$ be a function whose domain is the set of positive integers, and for positive integers $a$, $b$ and $n$, if $a + b = 2^{n}$, then $f(a) + f(b) = n^2$. What is $f(2021)$?

I started by testing values for $a$, $b$ and $n$, with the hope of finding a pattern, but so far I can't say I've made any headway;

For $a=b=n=1,\ f(1)+f(1)=2^1, so f(1)=\frac{1^2}{2}=\frac12$

I realize that it becomes easier to find $f(a)$ if $a$=$b$

So $a = b = 2^{n-1}$ then $f(a) = \frac{n^2}{2}$ that is, $f(2^{n-1}) = \frac{n^2}{2}$

At this point, I cannot see where to move forward.

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    $\begingroup$ Please edit to include your efforts. A natural starting point might be to find $f(k)$ for small values of $k$. $\endgroup$
    – lulu
    Jun 24 at 16:38
  • $\begingroup$ Note: may be worth remarking that $f(k)$ doesn't seem to be an integer for some small $k$. Nothing wrong with that, I suppose, but is it what you intended? $\endgroup$
    – lulu
    Jun 24 at 16:41
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    $\begingroup$ please include you efforts $\endgroup$ Jun 24 at 16:58
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    $\begingroup$ In your $a=b=n=1$ example, I think you mean: $1+1=2 = 2^1$, so $f(1) + f(1) = 1^2 = 1$ and thus $f(1) = 0.5$. $\endgroup$
    – rogerl
    Jun 24 at 17:15
  • $\begingroup$ Note that $2021+27=2048=2^{11}$ thus you can obtain $f(2021)$ in terms of $f(27)$. Repeat the process till you get a known value $\endgroup$ Jun 24 at 17:24

1 Answer 1

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A nearby power of $2$ to $2021$ is $2048=2^{11}$. So you know that $f(2021)+f(27)=11^2$ since $2021+27=2048$. Great, but now we need to know $f(27).$ Well, a nearby power of $2$ to $27$ is $32=2^5.$ So then because $27+5=32$, we know that $f(27)+f(5)=5^2$. Continue in this way until you get down to some small powers of $2$, which you can directly compute. Hope this helps.

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  • $\begingroup$ $2021+27=2048$, but $2021+37=2058$. $\endgroup$
    – A.M.
    Jun 24 at 17:28
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    $\begingroup$ Yikes, excuse me. Editing appropriately now. $\endgroup$
    – mwalth
    Jun 24 at 17:28
  • $\begingroup$ If possible, could you list out that continuation? $\endgroup$ Jun 24 at 20:53
  • $\begingroup$ Sorry, just seeing this request now. We would next need to find $f(5)$. For this, note that $5+3=8=2^3$, so $f(5)+f(3)=3^2=9$. Now we need $f(3)$, so we use $3+1=2^2$, so $f(3)+f(1)=2^2=4$. A comment by rogerl points out that $f(1)=.5$. Now you can back solve all the way up to $f(2021)$. $\endgroup$
    – mwalth
    Jun 28 at 16:37

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