-1
$\begingroup$

For any positive integer $n$, let $f(n)$ be the number of subsets of ${1, 2, . . . , n}$ whose sum is equal to $n$. Does there exist infinitely many positive integers $m$ such that $f(m) = f(m + 1)$? (Note that each element in a subset must be distinct.)

This is a New Zealand MO question and in their solution, they state that, by Michael Albert, that if we make $S(n)$ the set that fulfills what the question asks for, then $S(n)→S(n+1)$ is considered as a 'map', then this map would be an injection (needs proof but easy) and not a surjection provided that $S(n + 1)$ contains a set whose largest and second largest elements differ by one.

They state that a proof is needed and I would appreciate it if anybody could explain this to me, with the proof, as this will help me to understand the question better.

FYI: I have added my solution below.

Thank you very much!

$\endgroup$
3
  • 2
    $\begingroup$ You neither define $S(n),$ nor tell us what the map is. $X\to Y$ doesn't define anything, it just means some map from$X$ to $Y.$ Any map. $\endgroup$ Commented Aug 11 at 1:35
  • $\begingroup$ Let S(n) is the set of such subsets that fufills what the question states. $\endgroup$ Commented Aug 11 at 1:39
  • 2
    $\begingroup$ Don't put it in comments, put it in the question. As written, your question is horribly incomplete. $\endgroup$ Commented Aug 11 at 3:48

1 Answer 1

0
$\begingroup$

Let $S(n)$ be the set of such subsets. Consider the $S(n)$ to $S(n + 1)$ that adds one to the largest element of each $A ∈ S(n)$.

For even $n = 2k ⩾ 2$ this is true since we can take ${k, k + 1} ∈ S(n + 1)$ and for odd $n = 2k + 1 ⩾ 5$ this is true since we can take ${1, k, k + 1}$. So for $n ⩾ 5$, we must have $f(n) < f(n + 1)$ and there do not exist infinitely many such pairs.

$\endgroup$
2
  • $\begingroup$ Something seems to be missing here. What is true? $\endgroup$
    – Servaes
    Commented Aug 11 at 10:49
  • $\begingroup$ If there exist infinitely many positive integers $m$ such that $f(m)=f(m+1)$. $\endgroup$ Commented Aug 11 at 22:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .