Expert: Vijilant - 2/16/2010

Question
hello, sir or madam, my name is john, down here in Texas.  been going to grad school about the last two years, and now am taking a math history class this spring.  the question posed to you is as follows:  "show that in a Pythagorean triple, if the largest term is divisible by four, the so are the other two terms".  i have been thinking about this problem since this past weekend, ad have not gotten to far on it.  any help would be greatly appreciated.

Answer
Hello John

Think about what the other sqares could be.  They must be both odd or both even.
If both odd, (2x+1)^2 + (2y+1)^2 = 4(x^2 + x + y^2 +y) + 2  = 4z + 2.  There are no squares of this form.  Squares are of the form 4z or 4z+1.
Now let's try both even, and none of the form 4k.  4x+2 and 4y+2.
(4x+2)^2 + (4y+2)^2 = 16(x^2 + y^2 + x + y) + 8 which is not a multiple of 16 as it should be as the square of a multiple of 4.
Now let's try one a multiple of 4, the other not.
(4x)^2 + (4y+2)^2 = 16(x^2 + y^2 + y) + 4.  Again this is not a multiple of 16.

This was a first principles approach.  If you read further into Pythag triples you will find a simple rule to find if an integer n can be the hypotenuse.
Write down the prime factorisation of n.  Strike out all even powers, so you are left with a product of powers to the power 1.  If all of these are of the form 4k+1, it is possible.

We are now in a pythagorean free zone until 2017.
2010 has 2^1, 2011 is prime of form 4k-1, 2012 has prime 503 of form 4k-1.
2013 has 3^1, 2014 has 2^1, 2015 has 31^1, 2016 has 2^5, then at long last 2017 which is a prime of the form 4k+1.  Challenge: find the others in the triple starting 2017, and then work on the next ten years.  That should make a good paper to put before your fellow students.
(and amaze your professor).

Best wishes

vijilant (male of 71, so not pythagorean this year)