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# 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.

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