 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sum the integers, get squares? (Posted on 2007-04-05) Can the sum of any 2112 consecutive positive integers be a perfect square?

 See The Solution Submitted by K Sengupta Rating: 2.0000 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Different method | Comment 2 of 5 | Let x be the first positive integer in the sequence, so the sum of the 2112 consecutive positive integers would be x + (x + 1) + (x + 2) + ... (x + 2111).

This is equal to 2112x + (2111)(2112)/2, or simplified,

1056(2x + 2111).

We know that 1056 is not a perfect square, so 2x + 2111 would have to be equal to ��1056 for the whole expression to be a perfect square.  If you set 2x + 2111 = ��1056, you get

x = -1055.5

Since this is not a positive integer, we can conclude that the sum of any 2112 consecutive positive integers can not be a perfect square.

 Posted by Marc on 2007-04-05 10:09:15 Please log in:

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