I am Niranjan Srinivas from India. I am in Class X and I am also reading Higher Mathematics in an Olympiad Camp organised by the National Board of Higher Mathematics, India.
I would like to know the solution to this problem.

Prove that if n is a non-negative integer,it can be UNIQUELY represented in the form {( x + y)² + 3x + y }/2.

For this, I simplified the expression to
{(x + y)² + 2x + (x + y)}/2
= {(x + y)(x + y +1) + 2x}/2
= (x + y)(x + y +1)/2 + x
Since x + y and x + y + 1 are consecutive, one of them is even and thus
(x + y) (x + y + 1)/2 becomes an integer. Consequently, the expression is an integer, as an integer plus x (another integer) is always an integer.
But I was not able to establish the uniqueness.How should it be done ?

We need to know what values x,y are allowed to take. Are they integers? Do they have to be positive?

x and y are NON NEGATIVE integers.

OK, so let z = x + y. Now we want to show that every integer can be uniquely expressed as:

z(z+1)/2 + x

where x, z are integers with 0 £ x £ z.

Suppose z is fixed. What is the range of values that:

z(z+1)/2 + x

can take by varying x?

Well the lowest value it can take is:

z(z+1)/2

(as x is non-negative)

and the highest it can take is:

z(z+1)/2 + z = (z² + z + 2z)/2 = (z+1)(z+2)/2 - 1

It can clearly take all values in between as well by varying x (and it does so once only as the expression is linear in x). So for a fixed z the following numbers are available:

z(z+1)/2, z(z+1)/2 + 1, z(z+1)/2 + 2, ..., (z+1)(z+2)/2 - 1

exactly once.

It should now be clear why each number is uniquely expressible...

Once we've got to n = z(z+1)/2 + x, here's an alternative way of looking at it.

You will probably recognise z(z+1)/2 as the sum of the first z positive integers, or the zth triangular number. So n is the sum of a triangular number plus x.

Let's suppose there's more than one way of doing this. The obvious one is to take the highest triangular number smaller than n, k(k+1)/2, and then x will be the difference between this and n.

Since x cannot be negative, any other way must involve taking a smaller triangular number. The next largest is (k-1)k/2 = k(k+1)2-k. But this would mean that x would need to be whatever it was before, plus k. Since x < = k-1, this is impossible.

So the values of z and x are unique.