Prove that a=1 if 4(a^n+1) is a cube for all n (Iran 2008)

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sororak

337 posts

sororak

#1 h Sep 22, 2010, 7:13 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

$\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$ . We know that for every $n\in\mathbb{N}$ , $4(a^n+1)$ is a perfect cube. Prove that $a=1$ .

This post has been edited 1 time. Last edited by Amir Hossein, Mar 21, 2016, 7:48 PM

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jgnr

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jgnr

#2 h Sep 23, 2010, 8:27 AM • 15 Y

Y by Amir Hossein, GoJensenOrGoHome, kun1417, cyborg_w, siddigss, InCtrl, Hascotch, Ankoganit, aourkorzs, Adventure10, Mango247, GMMeowChand, and 3 other users

$4(a^3+1)$ and $4(a^9+1)$ are both perfect cubes, hence their quotient $a^6-a^3+1$ is also a perfect cube. For $a>1$ , $a^6-a^3+1<(a^2)^3$ , hence $a^6-a^3+1\le(a^2-1)^3$ , $3a^4-a^3-3a^2+2\le0$ , $(a-1)(3a^3-2a^2-1)+1\le0$ , a contradiction. Thus $a=1$ .

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amplreneo

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amplreneo

#3 h Nov 15, 2015, 3:43 PM • 3 Y

Y by Cosmobius, Batapan, Adventure10

Does anyone have a solution with LTE?

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buzzychaoz

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buzzychaoz

#5 h Nov 20, 2015, 6:37 AM • 7 Y

Y by wwwrqnojcm, cyborg_w, mamkin_botar, Ngankaka, Cosmobius, Numbertheorydog, Adventure10

Suppose $a\neq 1$ . If $a+1$ is not a power of $2$ , let $p$ be an odd prime dividing $a+1$ . Then by LTE $v_p(a^n+1)=v_p(a+1)+v_p(n)$ is a multiple of 3 for all odd positive integer $n$ , clearly impossible.

If $a+1$ is a power of $2$ , Note that $v_2(4(a^3+1))=2+v_2(a+1)+v_2(3)=2+v_2(a+1)+v_2(1)=v_2(4(a+1))$ , but $a^3+1>a+1$ implies there is an odd prime $p$ dividing $a^3+1$ . Then by LTE $v_p(4(a^{3k}+1))=v_p(a^3+1)+v_p(k)$ is a multiple of $3$ for all odd integers $k$ , thus contradiction and $a=1$ .

This post has been edited 2 times. Last edited by buzzychaoz, Nov 20, 2015, 9:16 AM
Reason: typo

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Takeya.O

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Takeya.O

#7 h Sep 6, 2016, 2:01 AM • 2 Y

Y by Adventure10, Mango247

I answer based on buzzychaoz's solution.

We suppose that $a>1$ .Obviously $2|a+1$ .By Zsigmondy's theorem,
∃ $p$ (prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$ .By LTE $\forall k$ (odd),
$v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$ .
$\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$
This is apparently absurd. $\blacksquare$

This post has been edited 1 time. Last edited by Takeya.O, Sep 6, 2016, 2:03 AM
Reason: Latex miss

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fighter

507 posts

fighter

#8 h Sep 6, 2016, 7:03 AM • 1 Y

Y by Adventure10

I have a same solution as jgnr also what is Zsigmondy's theorem

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Takeya.O

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Takeya.O

#9 h Sep 6, 2016, 7:15 AM • 2 Y

Y by Congruentisogonal44, Adventure10

@fighter
Brother! You should study Zsigmondy's theorem.This theorem kills many olympiad problem(number theory) instantly.

See here
http://www.math.ust.hk/excalibur/v16_n4.pdf

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fighter

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fighter

#10 h Sep 6, 2016, 9:40 AM • 2 Y

Y by Adventure10, Mango247

thank you

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math-helli

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math-helli

#11 h Sep 7, 2016, 5:32 PM • 1 Y

Y by Adventure10

A Problem by DR. M. Jamaali

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Takeya.O

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Takeya.O

#12 h Sep 8, 2016, 1:58 AM • 2 Y

Y by Adventure10, Mango247

@math-helli
Thanks,I see.

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buiminhducduc

1 post

buiminhducduc

#13 h Feb 28, 2017, 2:24 PM • 1 Y

Y by Adventure10

Takeya.O wrote:

I answer based on buzzychaoz's solution.

We suppose that $a>1$ .Obviously $2|a+1$ .By Zsigmondy's theorem,
∃ $p$ (prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$ .By LTE $\forall k$ (odd),
$v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$ .
$\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$
This is apparently absurd. $\blacksquare$

why n=3k?

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Takeya.O

769 posts

Takeya.O

#14 h Dec 11, 2017, 9:58 AM • 2 Y

Y by Adventure10, Mango247

buiminhducduc wrote:

Takeya.O wrote:

I answer based on buzzychaoz's solution.

We suppose that $a>1$ .Obviously $2|a+1$ .By Zsigmondy's theorem,
∃ $p$ (prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$ .By LTE $\forall k$ (odd),
$v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$ .
$\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$
This is apparently absurd. $\blacksquare$

why n=3k?

Now, natural number $n$ is arbitrary

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Αρχιμήδης 6

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Αρχιμήδης 6

#16 h Jan 23, 2018, 12:24 PM • 2 Y

Y by Adventure10, Mango247

Previous solutions are simpler but i will try to prove it with only one statement. $n=18$ so its going to be little harder.

I will prove that the only natural solution for $n=18$ is $a=1$ .

On diophantine equation $4(a^{18}+1)=x^3$

$4(a^{18}+1)=x^3$ so set $x=2y$

$a^{18}+1=2y^3$ so

$(a^6+1)(a^{12}-a^6+1)=2y^3$

obviously $a^{18}-a^6+1$ is odd so $a^6+1$ even. Also $(a^6+1,a^{12}-a^6+1)=3$ if $a^6=2mod3$ . If $a^6=0,1mod3$ then $(a^6+1,a^{12}-a^6+1)=1$ so we have 2 cases:

First case $(a^6+1,a^{12}-a^6+1)=1$ so

$a^6+1=2k^3$
$a^{12}-a^6+1=l^3$
$(k,l)=1$ , $(k,l)=y$

From second equation obiously $a$ bounded and finally we get $a=1$

Second case $(a^6+1,a^{12}-a^6+1)=3$ at this case notice that $a^{12}-a^6+1$ is never divisible by $9$ but divisible by $3$ so we get

$a^6+1=2*9k^3$
$a^{12}-a^6+1=3l^3$
$(k,l)=1$ , $kl=y$

Now we see that the equation $a^6+1=2*9k^3$ has no solutions $mod3$ .

So the only natural solutions of diophantine $4(a^{18}+1)=x^3$ are $(a,x)=(1,2)$

P.s

Can you see why i tried to solve this problem for $n=18$ ?

This post has been edited 2 times. Last edited by Αρχιμήδης 6, Jan 23, 2018, 12:46 PM
Reason: FIXED SOME TYPOS

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xdiegolazarox

42 posts

xdiegolazarox

#18 h Jan 23, 2018, 4:07 PM • 1 Y

Y by Adventure10

if we take $n=3k$ , we will have:

$(-a^{k})^{3}- 2(-y)^{3}=1$ for some integer $y$ , and by Skolem's theorem, $a$ must be $1$

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e_plus_pi

756 posts

e_plus_pi

#19 h May 28, 2018, 5:39 AM • 3 Y

Y by Satops, Adventure10, Mango247

My Solution

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grupyorum

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grupyorum

#20 h Jun 15, 2019, 3:06 AM • 2 Y

Y by Adventure10, Mango247

Here is a quite different solution than all of the above. Clearly, the thesis of the problem implies that there exists a sequence $(k_n)_{n=1}^\infty$ of positive integers, such that $a^n+1 = 2k_n^3$ holds. Now, $a^{n+3}+1 = 2k_{n+3}^3$ implies, $\frac{a^3-1}{2}=(ak_n)^3 - k_{n+3}^3$ . Now, if $a>1$ , then the sequence $k_n$ is strictly increasing (to infinity). Moreover, $a^2k_n^2 + ak_n k_{n+3}+k_{n+3}^2\mid \frac{a^3-1}{2}$ also holds. Thus, if $a>1$ , this divisibility implies a contradiction: $\frac{a^3-1}{2}$ is a fixed positive number, whereas $a^2k_n^2 + ak_n k_{n+3}+k_{n+3}^2\to+\infty$ .

This post has been edited 1 time. Last edited by grupyorum, Jun 15, 2019, 3:21 AM
Reason: typo fix.

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djmathman

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djmathman

#21 h Jun 9, 2020, 7:46 PM • 2 Y

Y by busy-beaver, Jack_w

I hope this works, because this LTE solution seems shorter than the others posted....

If $a^2 + 1$ has an odd prime factor, say $p$ , then LTE yields
$\nu_p(a^{2k} + 1) = \nu_p(a^2 + 1) + \nu_p(k)$ whenever $k$ is odd; this is clearly not always divisible by $3$ . Thus $a^2 + 1$ is a power of $2$ . But it's also congruent to $2\pmod 4$ , so $a = 1$ .

This post has been edited 2 times. Last edited by djmathman, Jun 10, 2020, 1:58 AM

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busy-beaver

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busy-beaver

#23 h Dec 11, 2024, 10:51 PM

Y by

Claim $(1)$ : $a^3 + 1$ is a power of two.
Proof:
SFTSOC that some odd $p \in \mathbb{P}$ divides $a^3 + 1$ . Then we also have $p \nmid a^3, 1$ and so (by LTE) for any odd $n \in \mathbb{Z}^+$ , $\nu_p\left(4\left(a^{3n} + 1^{3n}\right)\right) = \nu_p(4) + \nu_p\left(\left(a^3\right)^n + 1^n\right) = \nu_p\left(a^3 + 1\right) + \nu_p(n)$ Clearly there's some odd $n \in \mathbb{Z}^+$ for which the RHS is not divisible by $3$ , but then $4(a^n + 1)$ isn't a perfect cube, a contradiction. So, $a^3 + 1 > 0$ is not divisible by any odd prime, which implies it is a power of two.

Claim: $a < 3 \qquad (2)$ Proof: SFTSOC that $a \geq 3$ . Since $a^3 + 1 = (a + 1)(a^2 - a + 1)$ , by $(1)$ both $a + 1$ and $a^2 - a + 1$ are powers of two. Together with the fact that $a^2 - a + 1 > a + 1$ (because $a(a - 2) > 0$ ), this means we can write $a^2 - a + 1 = 2^i(a + 1)$ for some $i \in \mathbb{Z}^+$ . Taking the above equation modulo $a$ gives $2^i \equiv 1 \pmod{a}$ , and in particular $2^i > a$ (as $a > 1$ and $i > 0$ ). But then $2^i(a + 1) > a^2 + a > a^2 - a + 1$ , a contradiction.

Claim: $a = 1 \qquad (3)$ Proof:
If $a = 2$ , then $4(a^1 + 1) = 12$ is not a perfect cube, a contradiction. So, $a \neq 2$ , which means by $(2)$ that $a = 1$ , as desired.

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EaZ_Shadow

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EaZ_Shadow

#24 h Dec 11, 2024, 11:38 PM • 1 Y

Y by WHO_LET_ME_COOK