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Curious Consecutive Conundrum (Posted on 2008-04-08) Difficulty: 4 of 5
L and P are positive integers that satisfy this equation:

(L+1)3 L3 = P2

For example, 83 - 73 = 132; 1053 - 1043 = 1812, and so on.

Prove that P is always expressible as the sum of squares of two consecutive positive integers.

(For example, 13 = 22 + 32; 181 = 92 + 102, and so on.)

  Submitted by K Sengupta    
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Solution: (Hide)
By the given conditions, we have:

(L+1)3 - L3 = 3L2 + 3L +1 = P2

or, 12L2 + 12L + 4 = 4P2

or, 3(2L+1)2 = (2P+1)(2P-1)

Since each of 2P+1 and 2P-1 are odd, while (2P+1)-(2P-1) = 2, it follows that gcd(2P+1, 2P-1) = 1

Thus, for some integers x and y, it follows that:

(2P-1, 2P+1) = (3x2, y2), or: (x2, 3y2)

If the former case is true, then y2 = 3x2 + 2, so that:

y2(mod 3) = 2, which is a contradiction, since y2 (mod 3) must be congruent to either 0 or 1.

If the latter case is true, then:

x2 = 3y2 - 2

or, x2 (mod 3) = 1, which is indeed valid.

Again, x2 = 2P-1, so that x is odd. Thus, x = 2z+1, for some integer z

Accordingly, (2z+1)2 = 2P - 1

Hence: 4z2 + 4z + 1 = 2P-1

or, P = 2z2 + 2z + 1 = z2 + (z+1)2

Consequently, P is always expressible as the sum of squares of two consecutive positive integers.

-------------------------------------------------

Alternate Method:

(L+1)3 - L3 = P2......(i)

or, 3(2L+1)2 = (2P+1)(2P-1)......(ii)

From (i), it follows that P is odd.

Now, gcd(2P-1, 2P+1) = 1, as proved earlier, so in terms of (ii), it follows that at least one of 2P+1 and 2P-1 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, (2P-1) is a perfect square. Since 2P-1 is odd, it follows that:

2P - 1 = (2z + 1)2, whenever z is an integer.

or, P = z2 + (z+1)2

Consequently, P is always expressible as the sum of squares of two consecutive positive integers.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Hints/Tipsre: need a hintK Sengupta2008-07-23 16:11:47
need a hintPraneeth2008-07-23 03:40:32
Some Thoughts AND the n-th member is:Ady TZIDON2008-04-10 17:01:24
re: What value for L?Brian Smith2008-04-10 00:15:01
re: What value for L?brianjn2008-04-09 03:44:49
What value for L?brianjn2008-04-09 03:36:06
Some Thoughtsfirst thoughtsFrankM2008-04-08 22:58:08
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