1 + 2 + 3 + ... + n = n(n + 1)/2

And adding the successive odd numbers, we get

1 + 3 + 5 + ... + (2n - 1) = n^2

And if we sum the series of even numbers

2 + 4 + ... + 2n = n(n + 1)

8 * n(n + 1)/2 + 1 = 4n^2 + 4n + 1 = (2n + 1)^2

( 8 times any triangular number + 1 makes a square )The difference between the squares of any two consecutive triangular numbers is always

a cube:

http://mathworld.wolfram.com/TriangularNumber.html

The oblong number is double of a triangular number.

The successive oblong numbers are:

2*3 = 6, 3*4 = 12, 4*5 = 20, ... n(n + 1), ...

When is the Product of Two Oblong Numbers Another Oblong?

http://mathworld.wolfram.com/TriangularNumber.html

The oblong number is double of a triangular number.

The successive oblong numbers are:

2*3 = 6, 3*4 = 12, 4*5 = 20, ... n(n + 1), ...

When is the Product of Two Oblong Numbers Another Oblong?

Question :

Show that Gauss's discovery that every number is the sum of three or fewer triangular numbers implies that every number of the form 8k + 3 can be expressed as the sum of three odd squares

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