Is the following true, for each positive integer ## k ##?

[Math100]

Homework Statement

Is the following true, for each positive integer $k$?
$5\mid \sigma_{1} (5k+4)$
Briefly justify your answer.

Relevant Equations

None.

Let $5k+4=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}$.
Then $5\equiv 0\pmod {5}$ and $5k+4\equiv 4\pmod {5}$.
Thus $p_{i}^{k_{i}}\not \equiv 0\pmod {5}$ for $i=1, 2,..., s$.
Suppose all $p_{i}^{k_{i}}\equiv 1\pmod {5}$.
Then $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {5}$.
Since $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 4\pmod {5}$,
it follows that $\exists$ one $p_{i}$ satisfying $p_{i}^{k_{i}}\equiv 4\pmod {5}$.
This means $p_{i}\equiv 4\pmod {5}$.
Observe that $p_{i}^{2}\equiv 16\equiv 1\pmod {5}$ and $p_{i}^{3}\equiv 4\pmod {5}$.
If $p_{i}^{r}\equiv 4\pmod {5}$, then $r$ must be odd.
This implies $p_{i}^{k_{i}}\equiv 4\pmod {5}$ where $k_{i}$ is odd.
Now we have
\begin{align*}
&\sigma_{1} (p_{i}^{k_{i}})=p_{i}^{k_{i}}+p_{i}^{k_{i}-1}+\dotsb +p_{i}+1\\
&\equiv (4+1+\dotsb +4+1)\pmod {5}\\
&\equiv 0\pmod {5},\\
\end{align*}
because $p_{i}^{r}\equiv 4\pmod {5}$ if $r$ is odd and $p_{i}^{r}\equiv 1\pmod {5}$ if $r$ is even.
Thus
\begin{align*}
&5\mid \sigma_{1} (p_{i}^{k_{i}})\implies 5\mid [\sigma_{1} (p_{1}^{k_{1}})\dotsb \sigma_{1} (p_{i}^{k_{i}})\dotsb \sigma_{1} (p_{s}^{k_{s}})]\\
&\implies \sigma_{1} (p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}).\\
\end{align*}
Therefore, $5\mid \sigma_{1} (5k+4)$ for each positive integer $k$. And the following is true.

 

[fresh_42]

Homework Statement:: Is the following true, for each positive integer $k$?
$5\mid \sigma_{1} (5k+4)$
Briefly justify your answer.
Relevant Equations:: None.

Let $5k+4=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}$.
Then $5\equiv 0\pmod {5}$ and $5k+4\equiv 4\pmod {5}$.
Thus $p_{i}^{k_{i}}\not \equiv 0\pmod {5}$ for $i=1, 2,..., s$.
Suppose all $p_{i}^{k_{i}}\equiv 1\pmod {5}$.
Then $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {5}$.
Since $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 4\pmod {5}$,
it follows that $\exists$ one $p_{i}$ satisfying $p_{i}^{k_{i}}\equiv 4\pmod {5}$.

What about $p_1,p_2\equiv 2\pmod{5}$?

 

[Math100]

What about $p_1,p_2\equiv 2\pmod{5}$?

I am not sure. How did you get $p_{1}, p_{2}\equiv 2\pmod {5}$?

 

[fresh_42]

You said, that there must be a prime factor congruent $4$ modulo $5$. However, you did not cover the case $2\cdot 2=4.$ Assume $5k+4=p_1^{k_1}\cdot\ldots\cdot p_r^{k_r}.$ What if $p_1^{k_1}\equiv 2\pmod{5}$ and $p_2^{k_2}\equiv 2\pmod{5}$? E.g. consider $n=14.$

 

[Math100]

You said, that there must be a prime factor congruent $4$ modulo $5$. However, you did not cover the case $2\cdot 2=4.$ Assume $5k+4=p_1^{k_1}\cdot\ldots\cdot p_r^{k_r}.$ What if $p_1^{k_1}\equiv 2\pmod{5}$ and $p_2^{k_2}\equiv 2\pmod{5}$? E.g. consider $n=14.$

If $p_{1}^{k_{1}}\equiv 2\pmod {5}, p_{2}^{k_{2}}\equiv 2\pmod {5}$, then $p_{1}^{k_{1}}p_{2}^{k_{2}}\equiv 4\pmod {5}.$ You said to consider $n=14$, but where does $n=14$ come from?

 

[fresh_42]

If $p_{1}^{k_{1}}\equiv 2\pmod {5}, p_{2}^{k_{2}}\equiv 2\pmod {5}$, then $p_{1}^{k_{1}}p_{2}^{k_{2}}\equiv 4\pmod {5}.$ You said to consider $n=14$, but where does $n=14$ come from?

It is an example of why your proof does not work. $14=2\cdot 5 +4$ but no prime divisor or its power is congruent $4$ modulo $5.$

 

[Math100]

It is an example of why your proof does not work. $14=2\cdot 5 +4$ but no prime divisor or its power is congruent $4$ modulo $5.$

Then how should I disprove this? By using one of the counterexample(s) you've provided above?

 

[fresh_42]

I haven't provided any counterexamples. I just said why your proof cannot work. Have you calculated $\sigma_1(14)?$

 

[Math100]

I haven't provided any counterexamples. I just said why your proof cannot work. Have you calculated $\sigma_1(14)?$

No. What's $\sigma_1(14)?$

 

[fresh_42]

No. What's $\sigma_1(14)?$

Do you know what $\sigma_1$ means?

 

[Math100]

Do you know what $\sigma_1$ means?

No, what does it mean?

 

[fresh_42]

No, what does it mean?

It is the function that adds all divisors. So $\sigma_1(14)=1+2+7+14=24$ and $5\,\nmid\, 24.$

 

[Math100]

It is the function that adds all divisors. So $\sigma_1(14)=1+2+7+14=24$ and $5\,\nmid\, 24.$

I thought it's only the $\sigma(n)$, not the $\sigma_{1}(n)$.

 

[Math100]

So $\sigma_{1}(n)$ is the sum of all n's divisors.

 

[fresh_42]

I thought it's only the $\sigma(n)$, not the $\sigma_{1}(n)$.

This would be a matter of convention. Your problem used the function $\sigma_1$ so you must know what it means. In general, we have $\sigma_k(n)=\sum_{d|n} d^k.$

So $\sigma_{1}(n)$ is the sum of all n's divisors.

Yes. For $k=1$ we get $\sigma_1(n)=\sum_{d|n} d^1=\sum_{d|n}.$

 

[Math100]

Now I know that $5\mid \sigma_{1} (5k+4)$ isn't true for each positive integer $k$, because we just tested $k=2$.

 

[fresh_42]

Now I know that $5\mid \sigma_{1} (5k+4)$ isn't true for each positive integer $k$, because we just tested $k=2$.

Yes. The statement is false.

 

[Math100]

This would be a matter of convention. Your problem used the function $\sigma_1$ so you must know what it means. In general, we have $\sigma_k(n)=\sum_{d|n} d^k.$

Yes. For $k=1$ we get $\sigma_1(n)=\sum_{d|n} d^1=\sum_{d|n}.$

Then how should I disprove this statement? By letting $k=2$ as a counterexample?

 

[fresh_42]

Then how should I disprove this statement? By letting $k=2$ as a counterexample?

Yes, for example. Maybe you can find a second one.

 

[Math100]

Disproof:

Here is a counterexample:
Let $k=6$.
Then $\sigma_{1} (34)=1+2+17+34=54$.
Note that $5\nmid 54$.
Therefore, $5\nmid \sigma_{1} (5k+4)$ for each positive integer $k$. And the following statement is false.