**Problem: ** Show that there are infinitely many triples of positive integers, such that .

**Discussion **

Suppose we have found one such triplet (x, y, z). Then . Multiply to both sides where a is an arbitrary integer.

Clearly we have

Hence if (x, y, z) is a triple then 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 . Then r must be of the form 12k. Finally, their sum must be . This r+1 must be divisible by 31.

Let and we get . 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.

Hence we use this to form an equation:

Hence we have found one such triple : (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..