# West Bengal RMO 2015 Problem 3 Solution - Triples of Positive Integers

The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 3 Solution has been written for RMO preparation series.

## 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, Bézout's Theorem, Euclidean Algorithm, Sum of powers of two
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course, Cheenta Math Olympiad Program, discuss these topics in the ‘Number Theory’ module.
• Book Suggestions: Elementary Number Theory by David Burton

### 7 comments on “West Bengal RMO 2015 Problem 3 Solution - Triples of Positive Integers”

1. Debjitdj says:

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.

2. […] Triples of Positive Integers (RMO 2015 Problem 3) […]

3. Sayak Chatterjee says:

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

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

4. […] Show that there are infinitely many triples of positive integers, such that SOLUTION: Here […]

5. […] Show that there are infinitely many triples of positive integers, such that SOLUTION: Here […]

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