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Curious Consecutive Conundrum (Posted on 20080408) 

L and P are positive integers that satisfy this equation:
(L+1)^{3} – L^{3} = P^{2}
For example, 8^{3}  7^{3} = 13^{2}; 105^{3}  104^{3} = 181^{2}, and so on.
Prove that P is always expressible as the sum of squares of two consecutive positive integers.
(For example, 13 = 2^{2} + 3^{2}; 181 = 9^{2} + 10^{2}, and so on.)

Submitted by K Sengupta

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Solution:

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By the given conditions, we have:
(L+1)^{3}  L^{3} = 3L^{2} + 3L +1 = P^{2}
or, 12L^{2} + 12L + 4 = 4P^{2}
or, 3(2L+1)^{2} = (2P+1)(2P1)
Since each of 2P+1 and 2P1 are odd, while (2P+1)(2P1) = 2, it follows that gcd(2P+1, 2P1) = 1
Thus, for some integers x and y, it follows that:
(2P1, 2P+1) = (3x^{2}, y^{2}), or: (x^{2}, 3y^{2})
If the former case is true, then y^{2} = 3x^{2} + 2, so that:
y^{2}(mod 3) = 2, which is a contradiction, since y^{2} (mod 3) must be congruent to either 0 or 1.
If the latter case is true, then:
x^{2} = 3y^{2}  2
or, x^{2} (mod 3) = 1, which is indeed valid.
Again, x^{2} = 2P1, so that x is odd. Thus, x = 2z+1, for some integer z
Accordingly, (2z+1)^{2} = 2P  1
Hence: 4z^{2} + 4z + 1 = 2P1
or, P = 2z^{2} + 2z + 1 = z^{2} + (z+1)^{2}
Consequently, P is always expressible as the sum of squares of two consecutive positive integers.

Alternate Method:
(L+1)^{3}  L^{3} = P^{2}......(i)
or, 3(2L+1)^{2} = (2P+1)(2P1)......(ii)
From (i), it follows that P is odd.
Now, gcd(2P1, 2P+1) = 1, as proved earlier, so in terms of (ii), it follows that at least one of 2P+1 and 2P1 is a perfect square.
Recalling that P is odd, it follows that: (2P+1)(Mod 4) = 3, so that 2P+1 cannot be a perfect square.
Thus, (2P1) is a perfect square. Since 2P1 is odd, it follows that:
2P  1 = (2z + 1)^{2}, whenever z is an integer.
or, P = z^{2} + (z+1)^{2}
Consequently, P is always expressible as the sum of squares of two consecutive positive integers.

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