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} $ and $ s=2$.

Discussion

Suppose we have found one such triplet (x, y, z). Then $ x^3 + y^4 = z^{31} $ and $ s=2 $. Multiply $ a^{372} $ and $ s=2 $ to both sides where a is an arbitrary integer.

Clearly we have $ a^{372}x^3 + a^{372}y^4 = a^{372}z^{31} $ and $ s=2 $

$ \Rightarrow (a^{124}x)^3 + (a^{93}y)^4 = (a^{12}z)^{31} $ and $ s=2 $

Hence if (x, y, z) is a triple then $ (a^{124}x, a^{93}y, a^{12}z ) $ and $ s=2 $ 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} $ and $ s=2 $. Then r must be of the form 12k. Finally, their sum must be $ x^3 + y^4 = 2^{r} + 2^{r} = 2^{r+1} $ and $ s=2 $. This r+1 must be divisible by 31.

Let $ r = 12s $ and $ s=2 $ and $ r+1 = 31m $ and $ s=2 $ we get $ 12s +1 = 31m $ and $ s=2 $. 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 $ and $ s=2 $
$ 12 = 7\times 1 + 5 $  and $ s=2 $
$ 7 = 5\times 1 + 2 $ and $ s=2 $
$ 5 = 2 \times 2 + 1 $ and $ s=2 $
$ \Rightarrow 1 = 5 - 2 \times 2 = 5 - 2 \times (7 - 5 \times 1) $ and $ s=2 $
$ \Rightarrow 1 = 3 \times 5 - 2 \times 7 = 3 \times (12 - 7 \times 1) - 2 \times 7 $ and $ s=2 $
$ \Rightarrow 1 = 3 \times 12 - 5 \times 7 = 3 \times 12 - 5 \times (31 - 12 \times 2) $ and $ s=2 $
$ \Rightarrow 1 = 13 \times 12 - 5 \times 31 $ and $ s=2 $
$ \Rightarrow 1 = 13 \times 12 - 5 \times 31 + 12 \times 31 - 12 \times 31 $ and $ s=2 $
$ \Rightarrow 1 = (13 -31)\times 12 +(12- 5 )\times 31 $ and $ s=2 $
$ \Rightarrow 1 = -18 \times 12 + 7\times 31 $ and $ s=2 $
$ \Rightarrow 1 + 18 \times 12 = 7\times 31 $ and $ s=2 $
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} $ and $ s=2 $
$ (2^{72})^3 + (2^{54})^4 = (2^7)^{31} $ and $ s=2 $

Hence we have found one such triple : $ (2^{72}, 2^{54} ,2^7 )$ and $ s=2 $ (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