Understand the problem

Show that for each non-negative integer $n$ there are unique non-negative integers $x$ and $y$ such that we have
Source of the problem
Germany MO, 2019, Problem 4
Number Theory
Difficulty Level
Suggested Book
Challenges and Thrills of Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

If you observe the expression \( \frac{(x+y)^2 + 3x + y}{2} \). It has two free variables. Now, if let’s play with the expression to make it a little more symmetric and easy to work with and make some meaning out of it. Try to remove the asymmetry of x and write it in a much more symmetric form in x and y.

Observe that \( \frac{(x+y)^2 + 3x + y}{2} = \frac{(x+y)^2 + (x+y)}{2} + x\) is the expression what we get after bringing in the symmetry. Now, factorize it and see what we are looking for is \( n = \frac{(x+y)(x+y+1)}{2} + x\). Can you guess anything about the expression \( \frac{(x+y)(x+y+1)}{2} \).

\( \frac{(k)(k+1)}{2} \) is the sum of the first k natural numbers. So, now the idea is that somehow you are taking the first k natural numbers and adding another number x to it to make any number. Can you get the final logic?  

Now, observe that sum of the first k numbers is increasing with k. Now, take any number say 17. Now observe that 17 lies between 15 and 21. This means that 17 = 15 + 2 = ( 1 +2 + 3 +4 + 5) + 2. So, this is the idea that k = x+y is the largest number such that n is greater than or equal to the ( 1 + 2 + … + k ) and observe that we may need something like 2 in case of n = 17. Call that x and obviously \( k \geq x \). Hence, define y = k – x. So, for n = 33 = ( 1 + 2 + … + 7 ) + 5. Hence k = x + y = 7, x = 5, y = 2. This is the whole idea. QED.

Watch video

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Geometry of AM GM Inequality

AM GM Inequality has a geometric interpretation. Watch the video discussion on it and try some hint problems to sharpen your skills.

Geometry of Cauchy Schwarz Inequality

Cauchy Schwarz Inequality is a powerful tool in Algebra. However it also has a geometric meaning. We provide video and problem sequence to explore that.

RMO 2019 Maharashtra and Goa Problem 2 Geometry

Understand the problemGiven a circle $latex \Gamma$, let $latex P$ be a point in its interior, and let $latex l$ be a line passing through $latex P$. Construct with proof using a ruler and compass, all circles which pass through $latex P$, are tangent to $latex...

RMO 2019 (Maharashtra Goa) Adding GCDs

Can you add GCDs? This problem from RMO 2019 (Maharashtra region) has a beautiful solution. We also give some bonus questions for you to try.

Number Theory, Ireland MO 2018, Problem 9

This problem in number theory is an elegant applications of the ideas of quadratic and cubic residues of a number. Try with our sequential hints.

Number Theory, France IMO TST 2012, Problem 3

This problem is an advanced number theory problem using the ideas of lifting the exponents. Try with our sequential hints.

Algebra, Austria MO 2016, Problem 4

This algebra problem is an elegant application of culminating the ideas of polynomials to give a simple proof of an inequality. Try with our sequential hints.

Number Theory, Cyprus IMO TST 2018, Problem 1

This problem is a beautiful and simple application of the ideas of inequality and bounds in number theory. Try with our sequential hints.

Number Theory, South Africa 2019, Problem 6

This problem in number theory is an elegant applciations of the modulo technique used in the diophantine equations. Try with our sequential hints

Number Theory, Korea Junior MO 2015, Problem 7

This problem in number theory is an elegant application of the ideas of the proof of infinitude of primes from Korea. Try with our sequential hints.