Sum of Powers of Three Equals a Cube?

Number Theory Level 4

$\Large{3^x + 3^y + 3^z + 3^t = n^3}$

How many non-ordered pairs of non-negative integers $(x,y,z,t,n)$ exists satisfying the above equation?

1 0 5 3 4

\large{\infty}

Data Insufficient. Can't be solved. 2

3 solutions

Satyajit Mohanty
Aug 7, 2015

For any non-negative integers $a,b$ , we have:

$(3^a + 3^b)^3 = (3^a)^3 + 3(3^a)^2 (3^b) + 3(3^a)(3^b)^2 + (3^b)^3$ $= 3^{3a} + 3^{2a+b+1} + 3^{a+2b+1} + 3^{3b}$ .

Thus for all $x=3a, y=2a+b+1, z=a+2b+1, t=3b, n= 3^a + 3^b$ , where $a,b$ are non-negative integers, the equation satisfies.

Thus, infinite solutions.

Think you made a small typo in line $1$ - the coefficient of $(3^a)(3^b)^2$ should be $3$ . Nice solution!

Ryan Tamburrino - 5 years, 10 months ago

Thanks for letting me know about the Typo

Satyajit Mohanty - 5 years, 10 months ago

Interesting.

Vishnu Bhagyanath - 5 years, 10 months ago

Haha! :D Yes it is.

Satyajit Mohanty - 5 years, 10 months ago

汶良林
Aug 8, 2015

$3^{1} + 3^{1} + 3^{0} + 3^{0} = 2^{3}$

$( 3^{1} + 3^{1} + 3^{0} + 3^{0} )×3^{3n} = 2^{3}×3^{3n}$

$3^{3n+1} + 3^{3n+1} + 3^{3n} + 3^{3n} = (2×3^{n})^{3}$

$Thus, infinite solutions.$

Mathh Mathh
Aug 7, 2015

$(x,y,z,t,n)=(3k,3k,3k+1,3k+1,2\cdot 3^k)$ with $k\ge 1$ gives infinitely many non-negative 5-tuples $(x,y,z,t,n)$ satisfying the equation.