Vtyjkyuk

The problem asks to prove:
$$2mathbbZ+1= exists ninmathbbN s.t. .$$
where $d(n)$ is the function that counts the number of positive divisors of $n$.

I see that if the prime factorization of $n=p_1^m_1dots p_k^m_k$, then $d(n)=prod_i=1^k(m_i+1)$, therefore reduce the problem to
$$2mathbbZ+1=exists m_1,dots,m_kinmathbbN , s.t. .$$
But I have no idea how to proceed, please help.

If
$$|x|=prod_i=1^kfrac2m_i+1m_i+1tag 1$$
then
$$|x|prod_i=1^k(m_i+1)=prod_i=1^k(2m_i+1)equiv 1pmod 2$$
hence $xnotequiv 0pmod 2$ so that $xequiv 1pmod 2$, that's $x$ is odd.

Conversely, given $x$ an odd integer, we have to show that there exists $m_iinBbb N$ satisfying $(1)$.
Write $|x|=2^kq-1$ with $ninBbb N$ and $2nmid q$ and let $a=2^k-1$.
If $m_i=(a|x|-1)/2^i$ then $m_iinBbb N$ for $1leq ileq k$ and
$$|x|=qprod_i=1^kfrac2m_i+1m_i+1$$
and since $q<|x|$ and $q$ is odd, the assertion follows by induction on $q$.

I get the value $a=2^k-1$ in this manner.

Consider the recurrence $m_i+1=m_i/2$ and impose $2m_1+1=a|x|$ and $m_k+1=aq$ so that
$$|x|=qprod_i=1^kfrac2m_i+1m_i+1$$
Then $m_1=2^k-1m_k$ from which we get
$$a=frac2^k-1x$$
Taking $q$ such that $|x|=2^kq-1$ gives $a=2^k-1$.

Below is a somewhat different approach.

First consider, for any integers $a,,b$, we have $a=frac abcdot b$. For $frac ab$ to be of the form $frac2m+1m+1$, we take $b=fraca+12$.

So we are led to consider the sequence $$a_0(a)=a,a_1(a)=fraca_0(a)+12,cdots,a_n+1(a)=fraca_n(a)+12,cdots.$$

Now an easy induction shows that $a_n(a)=fraca+1+2+cdots+2^n-12^n=fraca+2^n-12^n$.
Hence, if $a$ is divisible by $2^n-1$, say $a=(2^n-1)k$, then $a_n(a)=(2^n-1)frack+12^n$.

Then, for any odd number $a$, we have
$$a=frac(2^n-1)a2^n-1=Xcdotfraca_n2^n-1,$$
where $X=prod_i=0^n-1fraca_ia_i+1$, $a_i=a_i((2^n-1)a)$, and
the previous discussion shows that $fraca_n2^n-1=fraca+12^n$

We can choose $n$ to be the greatest $ell$ such that $2^ell$ divides $a+1$. Then $fraca+12^n$ is an odd integer, and we have reduced from $a$ to a smaller odd number $fraca+12^n$, so an induction concludes the proof.

Also, observe that if $a_n$ is an integer, then so is $a_n-1=2a_n-1$. This justifies that $X$ is really of the form we want.

This is a rewrite of the previous version, and appears more intuitive I hope.