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Consecutive cubes and squares


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Show that if the difference of the cubes of two consecutive integers is the square of an integer, then this integer is the sum of the squares of two consecutive integers.
(The smallest non-trivial example is: 83 − 73 = 169. This is the square of an integer, namely 13, which can be expressed as 22 + 32.)

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(x+1)3 - x3 = c²

3x(x+1) = c² - 1

3x(x+1) = (c+1)(c-1)

if c is sum of consecutive squares

c = y² + (y+1)²

c = 2y² + 2y + 1

Then,

3x(x+1) = (2y²+2y + 2)(2y²+2y)

Take out 2 and 2y common from the 2 expressions

3x(x+1) = 4y(y²+y + 1)(y+1)

y² + y +1 = y(y+1) + 1, then

3x(x+1) = 4y(y+1) [(y(y+1) + 1)]

Let, y(y+1) = a

Then

3x(x+1) = 4a(a+1)

Effectively then, it needs to be shown that for some "x", there exists an "a" such that the sum of first x digits is 4/3 times the sum of first a digits.

Need help to prove this!

You can go further will the above analysis that since a = y(y+1), it must be even; so let a = 2n

Then, 3x(x+1) = 8n(2n+1)

Therefore, either x or x+1 must be a multiple of 8

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Interesting fact:


Y2+(Y+1)2 becomes (X+1)2-X2 under the change of variable X = Y2+Y.
So, the sum of squares of two integers can be transformed into the difference of squares of two other integers.

The first six non-trivial examples for the problem at hand are:
83-73 = 169 = 132 = (72-62)2
1053-1043 = 32761 = 1812 = (912 - 902)2
14563 - 14553 = 6355441 = 25212 = (12612 - 12602)2
202733 - 202723 = 1232922769 = 351132 = (175572 - 175562)2
2823603 - 2823593 = 239180661721 = 4890612 = (2445312 - 2445302)2
39327613 - 39327603 = 46399815451081 = 68117412 = (34058712 - 34058702)2
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