Divisible by $7^{n+1}$

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Divisible by $7^{n+1}$

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Mtlse

#1 h Aug 18, 2017, 1:11 PM • 4 Y

Y by rodamaral, Mathuzb, Adventure10, Mango247

Prove that for every non-negative $n$ , $3^{7^n}+5^{7^n}-1$ is divisible by $7^{n+1}$ .

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#2 h Aug 19, 2017, 3:13 PM • 2 Y

Y by Adventure10, Mango247

Induction

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#3 h Aug 20, 2017, 5:10 AM • 1 Y

Y by Adventure10

Could you explain me the detail of the induction proof？

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#4 h Aug 20, 2017, 3:45 PM • 7 Y

Y by claserken, doxuanlong15052000, naw.ngs, fungarwai, r_ef, Adventure10, Mango247

My solution: $3^{7^n}+5^{7^n}-1=$ -( $4^{7^n}+2^{7^n}+1)$
=- $\frac{8^{7^n}-1}{2^{7^n}-1}$ (mod $7^{n+1}$ )
Then use LTE

This post has been edited 1 time. Last edited by nmd27082001, Aug 20, 2017, 3:47 PM

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#5 h Aug 21, 2017, 6:12 AM • 1 Y

Y by Adventure10

can you show me about LTE

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#6 h Sep 1, 2017, 5:15 AM • 1 Y

Y by Adventure10

Mtlse wrote:

Prove that for every non-negative $n$ , $3^{7^n}+5^{7^n}-1$ is divisible by $7^{n+1}$ .

where did you find this problem?

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#7 h Sep 1, 2017, 5:46 AM • 2 Y

Y by Adventure10, Mango247

andreafort wrote:

can you show me about LTE

First prove that $7 \nmid 2^{7^{n}}-1$ , then we have $v_7(\frac{8^{7^n}-1}{2^{7^n}-1})=v_7(8^{7^{n}}-1)=v_7(8-1)+v_7(7^n)=n+1$ .

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#9 h Nov 6, 2017, 3:46 PM • 2 Y

Y by Adventure10, Mango247

Sorry where come from

$\frac{8^{7^n}-1}{2^{7^n}-1}$ ?

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#10 h Nov 6, 2017, 3:47 PM • 2 Y

Y by Adventure10, Mango247

And why and how [/quote] $3^{7^n}+5^{7^n}-1=$ -( $4^{7^n}+2^{7^n}+1)$
=- $\frac{8^{7^n}-1}{2^{7^n}-1}$ (mod $7^{n+1}$ ) [/quote]
Someone can show me the passages??

This post has been edited 2 times. Last edited by LittleKesha, Nov 6, 2017, 3:49 PM
Reason: error

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#11 h Nov 7, 2017, 1:43 AM • 1 Y

Y by Adventure10

LittleKesha wrote:

And why and how $3^{7^n}+5^{7^n}-1=$ -( $4^{7^n}+2^{7^n}+1)$

$3 \equiv 3 - 7 \equiv -4 \pmod{7}$ $3^{7^n} \equiv (-4)^{7^n} \pmod{7}$ $7 \mid 3^{7^n} - (-4)^{7^n}$ By LTE, $v_7(3^{7^n} - (-4)^{7^n}) = v_7(3 - (-4)) + v_7(7^n) = 1 + n$
Therefore, $7^{n+1} \mid 3^{7^n} - (-4)^{7^n}$ .
Likewise,
$5 \equiv 5 - 7 \equiv -2\pmod{7} \Rightarrow 7^{n+1} \mid 5^{7^n} - (-2)^{7^n}$ Back to the problem
$3^{7^a}+5^{7^a}-1 \equiv (-4)^{7^a}+(-2)^{7^a}-1 \equiv -(4^{7^a}+2^{7^a}+1) \pmod{7^a}$

LittleKesha wrote:

$\frac{8^{7^n}-1}{2^{7^n}-1}$ (mod $7^{n+1}$ )

This is pure algebra:
$x^3 - 1 = (x - 1)(x^2 + x + 1); \text{ make }x\text{ = }2^{7^n}$

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#12 h Nov 7, 2017, 7:53 AM • 2 Y

Y by Adventure10, Mango247

Why
$7 \nmid 2^{7^{n}}-1$ ?

This post has been edited 1 time. Last edited by LittleKesha, Nov 7, 2017, 7:53 AM
Reason: error

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#13 h Nov 7, 2017, 8:02 AM • 2 Y

Y by Adventure10, Mango247

I don't understand the first part of rodamaral's solution... is it only an example or a solution?

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#14 h Nov 7, 2017, 2:34 PM • 2 Y

Y by Adventure10, Mango247

LittleKesha wrote:

Why
$7 \nmid 2^{7^{n}}-1$ ?

$2^1 \equiv 2 \pmod{7}; 2^2 \equiv 4 \pmod{7}; 2^3 \equiv 1 \pmod{7}$ $ord_7(2) = 3 \Rightarrow (2^x \equiv 1 \pmod{7} \rightarrow 3 \mid x)$

LittleKesha wrote:

I don't understand the first part of rodamaral's solution... is it only an example or a solution?

It's just a detailed proof of a step that you asked.

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#15 h Nov 7, 2017, 2:58 PM • 2 Y

Y by Adventure10, Mango247

Induction.

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#16 h Nov 7, 2017, 8:37 PM • 2 Y

Y by Adventure10, Mango247

Ok thanks rodamaral!

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#17 h Nov 7, 2017, 8:37 PM • 1 Y

Y by Adventure10

Can you don2001 show me your resolution with the induction??

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#18 h Aug 16, 2018, 6:25 AM • 2 Y

Y by Adventure10, Mango247

Wow it's amazing

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TuZo

#19 h Aug 16, 2018, 1:55 PM • 1 Y

Y by Adventure10

I try to solve with induction, my work is the following, but I can not finalize, maybe somebody?
1) So we suppose that ${{3}^{{{7}^{n}}}}+{{5}^{{{7}^{n}}}}-1={{x}_{n}}\cdot {{7}^{n+1}}$
2) We must prove that $A={{3}^{{{7}^{n+1}}}}+{{5}^{{{7}^{n}}+1}}-1={{\left( {{3}^{{{7}^{n}}}} \right)}^{7}}+{{\left( {{5}^{{{7}^{n}}}} \right)}^{7}}-1={{\left( {{x}_{n}}\cdot {{7}^{n+1}}-{{5}^{{{7}^{n}}}}+1 \right)}^{7}}+{{\left( {{5}^{{{7}^{n}}}} \right)}^{7}}-1\vdots {{7}^{n+2}}$
3) We use this, from Newton binomials theoreme: ${{(a+b)}^{7}}=M{{a}^{2}}+6a{{b}^{6}}+{{b}^{7}}$
4) we get $A={{7}^{n+2}}m-{{({{5}^{{{7}^{m}}}}-1)}^{7}}+{{\left( {{5}^{{{7}^{m}}}} \right)}^{7}}-1\vdots {{7}^{n+2}}$
So we must prove that ${{({{5}^{{{7}^{m}}}}-1)}^{7}}-{{\left( {{5}^{{{7}^{m}}}} \right)}^{7}}+1\vdots {{7}^{n+2}}$ . How we prove this?

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