 # Understand the problem

Show that for each non-negative integer $n$ there are unique non-negative integers $x$ and $y$ such that we have $$n=\frac{(x+y)^2+3x+y}{2}.$$
##### Source of the problem
Germany MO, 2019, Problem 4
Number Theory
4/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

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.

# Connected Program at Cheenta

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

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.