2026/034) Prove that the number of integral solutions of the equation $x^3+y^4=z^{31}$ is infinite

Note: This is one method of solution. Other method exist 

Because the left had side has 2 terms and right had side has one and  

$2 * 2^p = 2^{p+1}$

If we can make 

  $x^3= y^4\cdots(1)$ 

Same as some power of 2 we have a solution

Because   $x^3$ is a power of 2 so  x has to be some power of 2 say 

$x=2^m\cdots(2)$ 

And similarly y has to be some power of two say 

 $x=2^n\cdots(3)$

From (1),(2), and (3) we have

$(2^m)^3 = (2^n)^4$

So 3m = 4n

As m and n are integers we have m must be divisible by 4 and n by 3 so 3m and 4n which are same by 12

So we have

$3m = 4n = 12k$

So

$m = 4k\cdots(4)$

$n= 3k\cdots(5)$

And from (3) and (4)

$x = 2^{4k}\cdots(6)$

$y =  2^{3k}\cdots(7)$

Putting in the given equation we get

$(2^{4k})^3 + (2^{3k})^4 = z^{31}$ 

Or $2^{12k} + 2^{12k} = z^{31}$

Or $2^{12k+1} = z^{31}$

Because LHS is a power of 2 so RHS is also a power of 2 so z has to be a power of 2 say 

$z= 2^t\cdots(8)$

Thus we get  $2^{12k+1} = 2^{31t}$

Or  $12k+1 = 31t\cdots(9)$

We can solve it using Extended Euclidean Algorithm to solve the same.

However as 12 and 31 are small numbers we can use the following approach as well

As $12 | 31t-1$ so $12 | 7t-1$ as 31 is 7 mod 12

By putting values of t from 0 to 11 we get (we need not put all values but upto the solution) and kowing that t is odd as t even shall make the number odd and not divisible by 12 we get t = 7.

Putting $t=7$ in (3) to get $k=18$

So one soultion is k =18, t = 7

As $12k+1 = 31t$ adding 12 * 31 a on both sides shall not change the values

So 12(k+31a) + 31(t +12a)

As one set of solution is (18,7) so parametric solution is  t = 7+12a, k= 18 + 31a

From (6) and (7) and (8) 

We get $x=2^{4(18+31a)}$,  $y=2^{3(18+31a)}$,  $z=2^{7+12a}$ 

This is a parametric solution and by varying a any whole number we can get any number of solution

Hence infinite number of solutions