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

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

What I have tried.

$$3^9+3^{12}+3^{15}+3^n$$

$$=3^9(757+3^{n-9})$$

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.

• How do you get 757? Are the powers in initial expression correctly written? – user376343 Aug 23 '19 at 7:00
• Use \$3^{15}\$ for $3^{15}$ and \$1 \pmod 3\$ for $1 \pmod 3$. Also, here is the MathJax reference. – Toby Mak Aug 23 '19 at 7:02
• 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 . – Jayant Jha Sep 4 '19 at 8:11
• Now posted to mathoverflow.net/questions/340131/… without notifying either site of the other posting. – Gerry Myerson Sep 8 '19 at 12:41
• 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. – 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 $$(a,n)=(10,14)$$

• Adding a link. Likely I am not the only reader who had not heard of Thue equations. +1, of course. – Jyrki Lahtonen Sep 2 '19 at 17:53
• 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. – 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 $$y^3 whenever $$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
• 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. – Jyrki Lahtonen Sep 2 '19 at 12:13
• Why $\epsilon\in\{0,1,2\}$? – Divya Prakash Sinha Sep 2 '19 at 17:28
• @DivyaPrakashSinha If $n-9=3k+\epsilon$ then $3^{n-9}=3^\epsilon\cdot(3^k)^3$. In other words, $y=3^k$. – 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 $$2 and $$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,
m=Mod(3,p);
h=znorder(m);
z=znlog(-757,m,h);
if(z,
c=iferr(chinese(Mod(1,2),Mod(z,h)), E, 0);
if(c,
m=Mod(3,p^2);
h=znorder(m);
z=znlog(-757,m,h);
if(z,
c=chinese(c,Mod(z,h));
m=Mod(3,p^3);
h=znorder(m);
z=znlog(-757,m,h);
if(z,
c=chinese(c,Mod(z,h));
j=c.mod; c=lift(c);
print("p = "p"    x = "c" + k*"j);
)
)
)
)
))


Output:

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,....