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Pythagorean Triples Divisible by 5

Date: 11/17/2000 at 17:44:40
From: Simon Essex
Subject: Do all integer right-angled triangles

I am trying to find the answer to whether all right-angled triangles 
with integer side lengths have a side with a length divisible by 5. I 
have no idea how to answer this.


Date: 11/17/2000 at 19:23:36
From: Doctor Mitteldorf
Subject: Re: Do all integer right-angled triangles 

I don't know what you mean by "I have no idea." Have you tried the 
first ten or twenty Pythagorean triples, and do they all have a side 
that's a multiple of 5?

If all of these test positive, the next step would be to investigate 
some formulas that generate Pythagorean triples, to see if there's 
some intrinsic reason why the formulas always have one number that's a 
multiple of 5. Do you know any such formulas?

For example the difference between n^2 and (n+1)^2 is 2n+1. This means 
that if you can find any odd number that's a square, you can set that 
equal to 2n+1 and then the three numbers are sqrt(2n+1), n^2 and 

Let me know if you find any counterexamples.

- Doctor Mitteldorf, The Math Forum   

Date: 11/18/2000 at 06:03:27
From: Doctor Mitteldorf
Subject: Re: Do all integer right-angled triangles

Dear Simon,

I've thought some more about your problem. First, I programmed a 
computer to look for Pythagorean triples up to a=10,000 and b=10,000, 
and hunt for ones in which there's no side divisible by 5. It seems 
that your suggestion is borne out: every single one had at least one 
side divisible by 5.

So that got me wondering the same thing you're probably wondering: why 
should that be so. So I started thinking in terms of "mod 5" 
arithmetic. If you're familiar with this, then you can work in that 
way, too; if not, here's a way to think about it.

Let's work in base 5. All numbers end in either 0, 1, 2, 3, or 4. The 
numbers divisible by 5 are the ones that end in 0. If you square a 
number that ends in 0, you get a number that ends in 0. If you square 
a number that ends in 1, you get a number ending in 1. Continuing:

     Number     Square
     ends in    ends in
        0          0
        1          1
        2          4
        3          4
        4          1

It's interesting to note that all squares end in 0, 1, or 4. The next 
step is to ask the same about the sum of two squares. What are the 
possible last digits for the sum of two squares? Which of these can be 
a square itself?

Follow this line of reasoning, and I think we'll have our answer to 
why it is that there always seems to be at least one side of the 
triangle that's divisible by 5.

- Doctor Mitteldorf, The Math Forum   
Associated Topics:
College Number Theory
High School Number Theory

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