x + 8 = 2x + 3, which gives the answer x = 5

What happens if we mix up the coefficients?

2x + 8 = 3x + 1 gives x = 7
3x + 2 = x + 8 gives x = 3

It is not hard to show that three natural number solutions are the most that you can get, starting with four integer coefficients.

So {1,2,3,8} generate the Triple {3,5,7}

Pythagorean triples are generated by a formula. Can anyone find a formula to generate all these Triples, and nothing but these Triples?

Sorry

sfwc
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From my Number Theory class that I took many years ago - the "reduced" Pythagorean triples are given by

( u^2 - v^2 , 2uv , u^2 + v^2 ) , where

1) 0 < v < u
2) (u,v) = 1
3) u + v is odd

Hope this helps.

Bractals

1. 8 cannot be a coefficient of x.

Given restriction 1, there are only 6 possibilities for mixing up the coefficients to have integer solution (-7, -5, -3, 3, 5, 7).

2. Coefficient of x on the equation side with 8 must be less than the coefficient of x at the other side of the equation.

Given restriction 2, the 3 negative solutions are eliminated and we are left with only 3 possibilities - not much of a mixing up.

To generate triples (3,5,7), the 4 integers for the equation must be {(n+m)q, (n+2m)q, (n+3m)q, (n+8m)q}, n > 0, m >= 1, q >= 1 are integers. For example, n = 15, m = 2, q = 2 leads to (34, 38, 42, 62). Observing the mixing rules 1 and 2:

34x + 62 = 38x + 42 => x = 5
34x + 62 = 42x + 38 => x = 3
38x + 62 = 42x + 34 => x = 7

If you want to generate a different triple, say (2k+1, 2k+3, 2k+5), using 4 integers for the equation (n+1, n+2, n+3, p), let's see what the 4th integer p must be. The generated solutions are

2k + 1 = {p - (n + 2)}/{(n + 3) - (n + 1)} = {p - (n + 2)}/2
2k + 3 = {p - (n + 3)}/{(n + 2) - (n + 1)} = p - (n + 3)
2k + 5 = {p - (n + 1)}/{(n + 3) - (n + 2)} = p - (n + 1)

p = 4k + 4 + n from (1)
p = 2k + 5 + n from (2) or (3)

4k + 4 + n = 2k + 5 + n
k = 1
p = n + 8

The consecutive odd triples you can generate are only (3, 5, 7). But suppose you do not require the first 3 numbers (of the 4 generators) to be consecutive (or multiples of consecutive numbers) and you want to generate the triple (5, 7, 9). Denote the 4 integers 0 < a₁ < a₂ < a₃ and p > 0.

5 = (p - a₂)/(a₃ - a₁)
7 = (p - a₃)/(a₂ - a₁)
9 = (p - a₁)/(a₃ - a₂)

This gives you 3 linear equations for 3 unknowns a₁, a₂, a₃

a₁ - 9a₂ + 9a₃ = p
-5a₁ + a₂ + 5a₃ = p
-7a₁ +7a₂ + a₃ = p

Triangularization of the matrix yields

+1, -9, +9, +p
-5, +1, +5, +p
-7, +7, +1, +p

+1, -9, +9, +p
+0, -44, +50, +6p
+0, 56, +64, +8p

+1, -9, +9, +p
+0, -7, +8, +p
+0, -22, +25, +3p

+1, -9, +9, +p
+0, -7, +8, +p
+0, +0, +1, +p

Since the 3 linear equations are linearly independent, the only solution (integer or not) is the trivial solution

a₃ = p
a₂ = (8p - p)/7 = p
a₁ = 9(p - p) + p = p

a₁ = a₂ = a₃ = p

and you cannot generate anything with that, because the generator numbers must be different. You can do this for any desired triple. The reason you can get a generator for the triple (3, 5, 7) is that the corresponding matrix happens to have linearly dependent rows:

+1, -7, +7, +p
-3, +1, +3, +p
-5, +5, +1, +p

+1, -7, +7, +p
+0, -20, +24, +4p
+0, -30, +36, +6p

+1, -7, +7, +p
+0, -5, +6, +p
+0, -5, +6, +p

so that you can find a non-trivial solution. Should you do this for a general triple of consecutive odd numbers (2k+1, 2k+3, 2k+5), the matrix triangularization would yield a nasty cubic equation for k with integer coefficients, for which you would have to find an integer root (if it existed) and I am not going to do that. However, I am at a complete loss about the purpose of this whole exercise.