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 T times T creates another T?? (Posted on 2015-01-11)
The squares of triangular numbers 1 and 6 are triangular numbers 1 and 36.

T1^2 = 1 * 1 = 1 = T1
T3^2 = 6 * 6 = 36 = T8

Are there additional triangular numbers whose squares form a triangular number?

 See The Solution Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

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 re(3): Solution | Comment 4 of 8 |
(In reply to re(2): Solution by Harry)

Harry:

I don't think you are quite out of the woods yet. There are two holes in the proof.

Problem 1:  It is not, for instance, correct that 14^2*15^2 can only be split into a product of two squares in one way.  It equals 7^2*30^2 and 42^2*5^2 and 1^2*210^2.  Also, note that 3^2*4^2 = 6^2*2^2, where both of the squares are even.  However, one of the terms must be roughly twice the other, so this can probably be dealt with.

Problem 2: You are assuming that either 2b and (b+1) are squares, or b and 2(b+1) are squares, but this is also not proved.  It does turn out to be true, however.  Because b and (b+1) are relatively prime, any prime factor greater than 2 is necessarily represented entirely in one term or another.  But that it is not necessarily the case with the factor of 2.  2b and (b+1), or b and 2(b+1), could conceivably both be 2 times a square.  But if 2b and (b+1) are both 2 times a square, then b and 2(b+1) are both squares. And if b and 2(b+1) are both 2 times a square, then 2b and (b+1) are both squares.  So it is always true that either 2b and (b+1) are squares, or b and 2(b+1) are squares.

So now I am agreeing with your conclusion, Harry.  I have fixed problem 2, and a little hand-waving can be used to fix problem 1.

Also, I feel something more elegant just out of reach.  Something about the problem 2 argument seems like it might get us directly to 'a' only being 1 or 3.

 Posted by Steve Herman on 2015-01-12 12:13:30
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