such that 
is an integer for every positive integer 
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For any positive integer 
denotes the sum of the positive integer divisors of 
Let 
be the least positive integer such that 
is divisible by 
for all positive integers Find the sum of the prime factors in the prime factorization of 
denotes the sum of the positive integer divisors of 
Let 
be the least positive integer such that 
is divisible by 
for all positive integers Find the sum of the prime factors in the prime factorization of 
This post has been edited 1 time. Last edited by IAmTheHazard, Mar 11, 2021, 4:58 PM
(mod 
)
(mod 
respectively. Answer format was a dead giveaway there were no repeated prime factors, and I also tested the 
case to be more sure of my answer. 
.
.
to get 
,
it just becomes 
,
.
we just get 
,
which divide 
so we can forget about those
cases. I even remembered they existed and messed up.
![\[\prod_{p^e\parallel a}\frac{p^{en+1}-1}{p-1}\equiv1\pmod{2021}\]](http://latex.artofproblemsolving.com/b/9/8/b9874b9531383862e69f7ae08b7c6569d2744d32.png)
for all 
,![\[\frac{p^{en+1}-1}{p-1}\equiv 1\pmod{2021}\]](http://latex.artofproblemsolving.com/e/a/4/ea4e38121ec7b8138a7d80435836cbe86dea0406.png)
for all 
,
.
always divides 
if and only if 
divides 
.
,
,
.
by Fermat's little theorem.
,
.
.
,
.
to hold: indeed, ![\[p^{en}\equiv\big(\text{const}\cdot q^k+1\big)^{en}\equiv en\cdot q^k+1\equiv1\pmod{q^{k+1}}.\]](http://latex.artofproblemsolving.com/5/f/f/5ff9fccca428cbb6d239defbc0a12509707150aa.png)
,
.
and multiplied a bunch of extra stuff to 
is divisible by 43 or 47 break the denominator of the fraction thingy
,![\[\frac{p^{en+1}-1}{p-1} \equiv p^{en} + p^{en-1} + \dots + 1 \equiv en + 1 \equiv 1 \pmod q,\]](http://latex.artofproblemsolving.com/2/5/1/251cd9c5b2b7e14d94478c7b71706698366f29cd.png)
i.e. 
.
and 
.
,
being relatively prime to 
to divide 
divide it is enough.
for all 
.
dividing 
for all integers 
,
and 
.
.
are coprime, then division is defined in mod 43 and we can just multiply it out. In that case, 
.
this is true, but we need to ensure this is true for all 
.
.
,
.
.
,
.
.
.
.
.
,
for any p. Thus, we just have 
and we do not need to modify 
.
,
,
is a prime. Note that 
. Our original modular congruence becomes 
This implies that 
.
and 
,
and 
.
or 
.
. We need 
,
.
.
.
.
.
then using LTE we find that 
.
then 
.
![\begin{align*}
&\text{We need}\; \sigma(a^n)-1 \equiv 0 \pmod {2021}.
\;\text{Checking for values:} \\
&[b]a=1:[/b] \sigma(1^n)-1=0 \;\text{so true for all n.}\\
&[b]a=p[/b], \text{(p prime)}: \;\sigma(p^n)-1=\frac{p(p^n-1)}{p-1} \equiv 0 \pmod{2021}.\\
&\text{Case I: p-1 has no 43s or 47s in its prime factorisation then it is easy to see that}\\&\; p(p^n-1)\equiv0 \pmod {2021} \; \text{we have to show when} \; p^n \equiv 1 \pmod{43} \; \text{and} \\ &\; p^n \equiv 1 \pmod{47},\; \text{n=42c and n=46d.} \;\text{So,} \;\text{n}=\text{LCM}(42,46)=42 \cdot 23.\\
&\text{Case II: p-1 has 43 or 47 in prime factorisation}, \\& \text{First we look if 43 is there}\; p-1=43^ke \\
&\Rightarrow p^n-1=43^{k+1}f\Rightarrow \boxed{p-1 \equiv 43^ke\pmod{43^{k+1}}} \\& \Rightarrow p^n \equiv (1+43^ke)^n \equiv n43^ke+1 \equiv 1\pmod{43^{k+1}} \\& \Rightarrow n43^ke \equiv 0 \pmod {43^{k+1}} \Rightarrow 43 \vert n.
\\& \text{Similarly we try for when 47 is there in prime factorisation of $p-1$ we get}\; 47\vert n\\
& \Rightarrow n=\operatorname{lcm}(42,43,46,47)\\
&[b]a is composite[/b]\\
&\text{Let}\; a=\prod_{i=1}^{u}p_i^{e_i} \;\text{Note that} \; \sigma(a) \text{is a multiplicative} \\
&\text{function:} \;\sigma(a)=\sigma\left(\prod_{i=1}^{u}p_i^{e_i}\right)=\prod_{i=1}^{u}\sigma\left(p_i^{e_i}\right), \; \text{as this product of} \text{is over all divisors of a.}\\
&\text{At}\; n=\operatorname{lcm}(42,43,46,47),\; \text{we have}\;
\sigma(a^n)-1=(\prod_{i=1}^{u}\sigma(p_i^{e_in}))-1 \equiv(\prod_{i=1}^{u}1)-1 \equiv 0\pmod{2021}.\\
&\text{The problems asks for the sum of the factors of n.}\\&n=\operatorname{lcm}(42,43,46,47)= 2\cdot3\cdot7\cdot23\cdot43\cdot47.\\
&\text{Therefore,}\; n=2+3+7+23+43+47=\boxed{\textbf{125}}
\end{align*}](http://latex.artofproblemsolving.com/5/4/7/547a348d027ce512ad7f990b986728af510df150.png)
.
,
,
we want![\[1 + p + p^2 + \dots + p^n = \dfrac{p^{n + 1} - 1}{p - 1} \equiv 1 \pmod{2021} \iff p^{n + 1} \equiv p \pmod{2021}. \]](http://latex.artofproblemsolving.com/2/b/d/2bdb5f41385bf6b15704e4f101996f43db0ddb45.png)
Let 
and 
be arbitrary primitive roots of 
and 
which forces 
.
,
or 
.![\[1 + p + p^2 + \dots + p^n \equiv n + 1 \equiv 1 \pmod{43} \iff 43 \mid n. \]](http://latex.artofproblemsolving.com/3/f/5/3f50c5e2c9ebdd0281b9fb28516b42e61d5f26d9.png)
A similar argument for 
.
.
,
![\[2021\phi(2021)\mid n\]](http://latex.artofproblemsolving.com/b/c/a/bcaa6c0974d6747490df4b51071be6417ccaa311.png)
is necessary,and then that it is sufficient,yielding 125.
.
.
.
,
so 
for all 
.
,
,
.
to satisfy this congruence for all 
,
.
.
,
,
