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March 13[edit]

Pythagorean triple properties[edit]

A few years ago, the Pythagorean triple article had properties about numbers related to the terms a, b, and c about divisibility by numbers greater than 5. Someone removed them as "unverified claim". Can anyone find a reliable source that reveals a pattern in statements of the following kind:

In a Pythagorean triple, exactly one of a group of n variables defined by a, b, and c is divisible by 2n-1, if 2n-1 is prime.

Examples:

Exactly one of a, b, is divisible by 3.
Exactly one of a, b, c is divisible by 5.
Exactly one of a, b, a+b, b-a is divisible by 7. (This was the first property that was removed.)

These properties get more complicated as the value of n gets larger. (Only values where 2n-1 is prime are allowed here.) For example, there will be 106 variables total that exactly one of is divisible by 211. Any thoughts on whether there are any sources revealing a formula for the general equation that the above statements are the first few examples of?? Georgia guy (talk) 20:05, 13 March 2015 (UTC)[reply]

I'm fairly sure that it's not too difficult to prove the statements you mentioned using mathematical induction on Euclid's formula. As for the pattern, I'm immediately struck by how this feels, to me, like the Fermat polygonal number theorem, but this is well outside my competence to comment. Also, I'd have thought on the basis of the negligible pattern we have to work with so far that I could construct 47 variables, of which exactly one is divisible by 211. :-) RomanSpa (talk) 00:55, 14 March 2015 (UTC)[reply]

The number of variables needed where exactly one is divisible by 211 is 106, not 47. The properties of the variables are:

The monomial variables a and b will always occur.
The monomial variable c will occur if and only if n is odd (but remember that only values of n where 2n-1 is prime are allowed)
Exactly one of the variables will be 2n-1 in the triple (3,4,5)
The monomial a is the only variable that will be 0 in the triple (0,1,1) (so we can rule out c-b for example)

Whenever a variable with at least one of a and b occurs, the same variable with a and b interchanged will also occur.

If a binomial variable occurs with a + sign, the same binomial variable with a minus sign will also occur (with absolute value added if necessary)

The following is what I know: (the definition of n here is that exactly one of the n variables is divisible by 2n-1; the domain of n here is the set of natural numbers where 2n-1 is prime)

n=2: a, b
n=3: a, b, c
n=4: a, b, a+b, b-a
n=6: a, b, (2a+b), |2a-b|, (2b+a), (2b-a) (important note: use the absolute value sign for variables where it would otherwise be allowed to be either positive or negative)
n=7: a, b, c, (2c+a), (2c-a), (2c+b), (2c-b)

Georgia guy (talk) 13:11, 14 March 2015 (UTC)[reply]

I think all these properties of the variables are confusing the issue. A more general statement is this: Let C be a nondegenerate quadratic curve in the projective plane over a finite field, and suppose the number of points on C is an even number 2n. Then there are n lines so that every point on C lies on exactly one line. To prove this, pair up the points in any way, say {p_i, q_i} for 1 = 1 to n. Now let line i be the line through p_i and q_i. Every point on C is on at least one line by definition, and a point being one more than one line would imply that the one of the lines intersects C in more than two points, which is impossible for a nondegenerate quadratic curve.

To see how to use this to generate a rule for p=11 (n=6), note that c=0 does not intersect the curve since 11 is not equal to 1 mod 4. So assuming c=1, the points on the curve are (a, b) = (±1, 0), (0, ±1), (±3, ±5), (±5, ±3), a total of 12 point. The are many ways of pairing up the points but one way that gives simple equations is {(±1, 0)}, {(0, ±1)}, {(3, 5), (5, 3)}, {(-3, 5), (-5, 3)}, {(3, -5), (5, -3)}, {(-3, -5), (-5, -3)}, giving the lines b=0, a=0, a+b=8c, -a+b=8c, a-b=8c, -a-b=8c. This in turn gives the rule, "Exact one of a, b, a+b+3c, -a+b+3c, a-b+3c, -a-b+3c is divisible by 11". (The triple (3, 4, 5) satisfies the third of these.) Another pairing is {(±1, 0)}, {(0, ±1)}, {(3, 5), (-3, -5)}, {(-3, 5), (3, -5)}, {(-5, 3), (5, -3)}, {(5, 3), (-5, -3)}, which gives the lines a=0, b=0, a±2b=0, b±2a=0, and the rule, "Exactly one of a, b, a+2b, a-2b, 2a+b, 2a-b is divisible by 11. (The triple (3, 4, 5) satisfies the third of these also. This is the n=6 case given above.)

The real problem then is to show a²+b²=c² is non-degenerate over any field of of order p. I haven't tried to do this yet though. --RDBury (talk) 16:18, 14 March 2015 (UTC)[reply]

I just checked this; it's true as long as characteristic is not 2. In char 2 a²+b²+c²=(a+b+c)² so the curve is degenerate. --RDBury (talk) 16:46, 14 March 2015 (UTC)[reply]

That is just like my case above when it comes to n=6. Can you try n=7 to see if my set of variables is correct??

For n=7, or p=13, the points on the curve are (2, 3, 0), (3, 2, 0), (±1, 0, 1), (0, ±1, 1), (±2, ±6, 1), (±6, ±2, 1) in projective coordinates. With the pairing {(2, 3, 0), (3, 2, 0)}, {(±1, 0, 1)}, {(0, ±1, 1)}, {(2, ±6, 1)}, {(-2, ±6, 1)}, {(±6, 2, 1)}, {(±6, -2, 1)} you get the variables you listed. You can also pair them as {(2, 3, 0), (3, 2, 0)}, {(±1, 0, 1)}, {(0, ±1, 1)}, {(2, 6, 1), (-2, -6, 1)}, {(2, -6, 1), (2, -6, 1)}, {(6, 2, 1), (-6, -2, 1)}, {(6, -2, 1), (-6, 2, 1)}, which would give the variables c, b, a, 3a-b, 3a+b, a-3b, and a+3b. It appears you can always have a set of expressions involving only a and b if n is even, and a set of variables with c and others involving only a and b if n is odd. --RDBury (talk) 17:04, 15 March 2015 (UTC)[reply]

That's a good hint. Georgia guy (talk) 19:34, 15 March 2015 (UTC)[reply]