Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.

What I have tried.



Let $757+3^{n-9}=a^3$

Taking modulo $3$:

$a^3\equiv 1 \pmod 3$

$\implies a\equiv 1 \pmod 3$

Also, $a^3>757>729=9^3$

$\therefore a =10+3k$

$a=10$ gives $n=14$

Now, how can I know if there is any other $n>14$ satisfying the given condition.

PS: Please not use computer programmes to answer. I want pure mathematical solution.

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    $\begingroup$ How do you get 757? Are the powers in initial expression correctly written? $\endgroup$ – user376343 Aug 23 '19 at 7:00
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    $\begingroup$ Use \$3^{15}\$ for $3^{15}$ and \$1 \pmod 3\$ for $1 \pmod 3$. Also, here is the MathJax reference. $\endgroup$ – Toby Mak Aug 23 '19 at 7:02
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    $\begingroup$ Note that $1000-757=243=3^5$. We can somehow use this by substituting $a$ by $10+3k$, cubing it and then subtracting the constants . $\endgroup$ – Jayant Jha Sep 4 '19 at 8:11
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    $\begingroup$ Now posted to mathoverflow.net/questions/340131/… without notifying either site of the other posting. $\endgroup$ – Gerry Myerson Sep 8 '19 at 12:41
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    $\begingroup$ As Gerry Myerson pointed out, it is reasonable to follow the recommendations explained on meta when cross-posting. {There is also a post with some guidelines about cross-posting on MathOverflow Meta.) Although it seems that the MO copy might get closed mathoverflow.net/review/close/120015 - so in this particular case it probably does not matter that much. $\endgroup$ – Martin Sleziak Sep 8 '19 at 12:46

Check manually if there exist any solution with $n \le 9$.

Now assume $n>9$, and look at the equation $3^m+757=a^3$ modulo 7 (with $m=n-9$):

$3^m+1 \equiv 3^m+757 \equiv a^3 \equiv \{0,-1,1\} \pmod 7$

$3^m \equiv \{0,-1,-2\} \pmod 7$

$n \equiv \{3,5\} \pmod 6$

If $m=6k+3$,

$ 757 = a^3-(3^{2k+1})^3 = (a-b)(a^2+ab+b^2)$

where $b=3^{2k+1}$. Factoring 757, you can check there is no solutions.

Else, if $m=6k+5$,

$757 = a^3-3^5(3^{2k})^3 = a^3-3^5b^3$

where $b=3^{2k}$. This is a Thue equation, effectively solvable:

using PARI/GP

tnf = thueinit(x^3-243)

thue(tnf, 757)

[[10, 1]]

you can check the only solution is $(a,b)=(10,1)$, hence $(a,n)=(10,14)$

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    $\begingroup$ Adding a link. Likely I am not the only reader who had not heard of Thue equations. +1, of course. $\endgroup$ – Jyrki Lahtonen Sep 2 '19 at 17:53
  • $\begingroup$ For the manual check note that $n\le9$ would have to be a multiple of $3$, else the expression has a wrong number vhf of powers of $3$ in its factorization. $\endgroup$ – Oscar Lanzi Sep 4 '19 at 10:02

Not a full solution, I am simply reducing the problem to that of listing integer points on two elliptic curves. IIRC this is implemented in some dedicated CAS, and therefore this gives us a route to a definite answer.

With small values of $n$ checked by brute force, we can cancel the factor $3^9$. We are thus left with the equation $$ 1+3^3+3^6+3^{n-9}=x^3\Longleftrightarrow 757+3^{n-9}=x^3.\qquad(*) $$ Depending on the residue class of $n$ modulo three we can write $3^{n-9}=3^\epsilon y^3$ with $\epsilon\in\{0,1,2\}$. This means that any integer solution of $(*)$ will give rise to an integer solution of one of the following Diophantine equations $$ \begin{aligned} x^3&=y^3+757,\\ x^3&=3y^3+757,\\ x^3&=9y^3+757. \end{aligned} $$ Each of these defines an elliptic curve. Those are known to have only finitely many integer points $(x,y)$, and (IIRC) algorithms for finding them exist (and are available in CAS's heavily used by number theorists).

Given such finite lists, we can quickly check whether $y$ can be a power of three in any of them.

Further remarks:

  • The first elliptic curve won't produce solutions. We have $$y^3<y^3+757<(y+1)^3$$ whenever $y>16$, and it is easy to check that the powers of three in this range won't give us any solutions.
  • Peter did an extensive computer verification for a largish range of values of $n$ (see the comments under main). So even an upper bound on the integer points (don't remember whether useful ones are known) will help us settle the main question
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  • $\begingroup$ This is mostly an oversized comment. Unlikely to satisfy the asker, or many readers of this question, including yours truly, even if it leads to an answer. $\endgroup$ – Jyrki Lahtonen Sep 2 '19 at 12:13
  • $\begingroup$ Why $\epsilon\in\{0,1,2\}$? $\endgroup$ – Divya Prakash Sinha Sep 2 '19 at 17:28
  • $\begingroup$ @DivyaPrakashSinha If $n-9=3k+\epsilon$ then $3^{n-9}=3^\epsilon\cdot(3^k)^3$. In other words, $y=3^k$. $\endgroup$ – Jyrki Lahtonen Sep 2 '19 at 17:50

Only long comment.

Let $n-9=x$, then we have equation $757+3^x=a^3$ for positive integers $a,x$.

$2\mid a$ and $a\equiv 1\pmod3$, then some prime $2<p\mid a$ and $p\equiv 2\pmod3$.

Find suitable primes $p$ and related $x$ using descrete logarithm, multiplicative order and CRT in pari/gp.

First, $x$ is odd, becose $8\mid a^3$, $ord_8(3)=2$ and $zlog_{3_{8}}(-757)=1$:

? m=Mod(3,8);
? h=znorder(m)
%1 = 2
? znlog(-757,m,h)
%2 = 1

gp-code for finding $p$ and $x$:

 forprime(p=5,100, if(p%3==2,
   c=iferr(chinese(Mod(1,2),Mod(z,h)), E, 0);
      j=c.mod; c=lift(c);
      print("p = "p"    x = "c" + k*"j);


p = 5    x = 5 + k*100
p = 23    x = 8169 + k*11638
p = 29    x = 11719 + k*23548
p = 47    x = 28121 + k*101614
p = 71    x = 332919 + k*352870

I.e. if $23\mid a$, then $x=8169 + k\cdot 11638$, where $k$ is 0,1,2,....

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