**Problem: ** Show that there are infinitely many triples \((x,y,z)\) of positive integers, such that \(x^3+y^4=z^{31} \).

**Discussion **

Suppose we have found one such triplet (x, y, z). Then \(x^3 + y^4 = z^{31} \). Multiply \(a^{372} \) to both sides where a is an arbitrary integer.

Clearly we have \(a^{372}x^3 + a^{372}y^4 = a^{372}z^{31} \)

\(\Rightarrow (a^{124}x)^3 + (a^{93}y)^4 = (a^{12}z)^{31} \)

Hence if (x, y, z) is a triple then \((a^{124}x, a^{93}y, a^{12}z ) \) is another such triple. Since a can be any arbitrary integer, hence we have found infinitely many such triplets ** provided we have found at least one **

To find one such triple, we use the following intuition: set x, y, z as some powers of 2 such that \(x^3 = y^4 = 2^{r} \). Then r must be of the form 12k. Finally, their sum must be \(x^3 + y^4 = 2^{r} + 2^{r} = 2^{r+1} \). This r+1 must be divisible by 31.

Let \(r = 12s \) and \(r+1 = 31m \) we get \(12s +1 = 31m \). Since 12 and 31 are coprime there is integer solution to this linear diophantine equation (by Bezoat’s theorem). We can solve this linear diophantine equation by euclidean algorithm.

\(31 = 12 \times 2 + 7 \)

\(12 = 7\times 1 + 5 \)

\(7 = 5\times 1 + 2 \)

\(5 = 2 \times 2 + 1 \)

\(\Rightarrow 1 = 5 – 2 \times 2 = 5 – 2 \times (7 – 5 \times 1) \)

\(\Rightarrow 1 = 3 \times 5 – 2 \times 7 = 3 \times (12 – 7 \times 1) – 2 \times 7 \)

\(\Rightarrow 1 = 3 \times 12 – 5 \times 7 = 3 \times 12 – 5 \times (31 – 12 \times 2) \)

\(\Rightarrow 1 = 13 \times 12 – 5 \times 31 \)

\(\Rightarrow 1 = 13 \times 12 – 5 \times 31 + 12 \times 31 – 12 \times 31 \)

\(\Rightarrow 1 = (13 -31)\times 12 +(12- 5 )\times 31 \)

\(\Rightarrow 1 = -18 \times 12 + 7\times 31 \)

\(\Rightarrow 1 + 18 \times 12 = 7\times 31 \)

Hence we use this to form an equation:

\(2^{18 \times 12} + 2^{18 \times 12} = 2^{216} + 2^{216} = 2^{217}=2^{7\times 31} \)

\((2^{72})^3 + (2^{54})^4 = (2^7)^{31} \)

Hence we have found one such triple : \((2^{72}, 2^{54} ,2^7 ) \) (from which we have shown earlier that infinitely more can be generated)

## Chatuspathi:

**What is this topic:**Number Theory**What are some of the associated concept:**Linear Diophantine Equation, Bezoat’s Theorem, Euclidean Algorithm, Sum of powers of two**Where can learn these topics:**Cheenta**Book Suggestions:**Elementary Number Theory by David Burton

I found, in some question papers, it is mentioned only ‘integers’ not ‘positive integers’. I know, that it should be ‘positive integers’, otherwise it will become a trivial problem by taking x=0.

That is right 🙂

But in original question (x,y,z) were said to be only integers; not necessarily positive.

Hmm… I think the intension was positive integer solution.. Otherwise the problem becomes trivial..