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Re: triangular numbers and dets
Posted:
Apr 12, 2012 2:37 PM
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> To get a=7 and b=33 did you do this by trial and
> error
> or by a direct method?
I actually wasted enough time to come up with a
parametric solution to the problem, expressing all
"primitive" solutions in terms of two integers m and n.
[by primitive meaning no common divisors between any
two of x,y,z]
I used x and y for variables, so I started with
[1] x^2 + x*y + y^2 = z^2.
For a primitive solution x,y cannot both be even, and
so I made the assumption that y is odd.
Then multiply [1] by 4 and rearrange to get
[2] 3y^2 = 4z^2 - (2x+y)^2,
[3] 3y^2 = (2z - 2x - y)*(2z + 2x + y).
From primitivity one can now show the two factors
on the right are relatively prime. (for the prime 2 this
is where the assumption that y is odd comes in).
So in the solution, one factor is 3 times a square,
while the other is a square. Call the squares m^2 and n^2.
After algebra on the two cases, we come up with two
somewhat different looking formula sets:
case 1: x = (m+n)*(n-3m)/4, y = m*n, z = (3m^2+n^2)/4.
case 2: x = (m-n)*(n+3m)/4, y = m*n, z = (3m^2+n^2)/4.
Note both cases have the same formula for y,z but,
supposing we want all of x,y,z positive,
case 1 requires n>3m while case 2 requires m>n.
To get integer solutions and make sure x,y,z is primitive,
I'm pretty sure we need that:
1: m and n are odd
2: The inequalities going with the cases
3: gcd(m,n) = 1.
4: n is not divisible by 3.
Anyway, plugging in various m,n satisfying these
conditions does give lots of solutions, and I think
gives them all.
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