One more rule about Pythagorean triples

Yet again, let's consider Pythagorean triples, sets of three positive integers a, b and c such that a² + b² = c².  Here are several examples where all three numbers are less than 100.

3² + 4² = 5²
12² + 5² = 13²
15² + 8² = 17²
7² + 24² = 25²
9² + 40² = 41²
20² + 21² = 29²
48² + 55² = 73²

We know that we can generate Pythagorean triples using any two positive integers j and k with the formulas, where j > k with

a = j² - k²    b = 2jk and   c = j² + k²

This means there must be infinitely many different Pythagorean triples. If you aren't sure, consider k=1 and we get

a = j² - 1   b = 2j and  c = j² + 1

For any j > 1, we will get a unique Pythagorean triple.

Here's a pattern I noticed. Every triple on our list above has one number that is divisible by 5. Let us start with a speculation.

Speculation: Every Pythagorean triple has at least one number that is divisible by 5.

We could create a triple where all three numbers are divisible by 5 just by taking any Pythagorean triple and multiplying all the numbers by 5.

Example: Because  3² + 4² = 5², 15² + 20² = 25², The original numbers are 9 +16 = 25, and if we multiply all the numbers by 5, all the squares are multiplied by 25 and we get 225 + 400 = 625.

That explains why the modifier "at least" is used in the speculation.

Let's find facts that will help us prove the speculation.

Fact: The last digit of a square is determined by the last digit of the original number.

Think about multiplying two numbers that have two digits together, like 72 × 72. 

__72
_×72
_144
504 
5184

The ones place is just 2 × 2 and the 70 isn't involved. This would be true no matter how many digits in a number.

Fact: The last digit of a perfect square can only be 0, 1, 4, 9 or 6.

We can prove this by looking at the list of perfect squares of the numbers 0 through 9 (0, 1, 4, 16, 25, 36, 49, 64, 81) and using the first fact from above that the last digit of a number determines the last digit of the square.

Fact: The remainder of a perfect square when divided by 5 can only be 0, 1 or 4.

Because 10 = 2 × 5, the remainder upon division by 5 will look a lot like the last digit, except 6 becomes 1, and 9 becomes 4.

Fact: If you want to make a sum of two numbers from the set {0, 1, 4} to add up to 0, 1 or 4, there are only a finite number of patterns and not all of them work.

0 + 0 = 0 works
0 + 1 = 1 works
0 + 4 = 4 works
1 + 0 = 1 works
1 + 1 = 2 doesn't work
1 + 4 = 5 doesn't work, except when we divide 5 by 5 we get remainder 0, so it does work for us.
4 + 0 = 4 works
4 + 1 = 5  doesn't work, except when we divide 5 by 5 we get remainder 0, so it does work for us.
4 + 4 = 8 doesn't

Fact: Every pattern that works has a 0 or 5 in it.

Fact: A remainder of 0 when dividing by five means the number is divisible by 5.

Fact: Because 5 is prime, getting a perfect square that is divisible by 5 means the original number is divisible by 5.

This means our speculation about the Pythagorean triples is in fact a pattern that is always true. We could call it a "theorem", but I'm going to save that word for

Fact: Every Pythagorean triple has at least one number that is divisible by 5.

Here are two other facts that can be proved using similar methods, which the reader can try to prove if interested.

Fact: Every Pythagorean triple has at least one number that is divisible by 3.

Fact: Every Pythagorean triple has at least one number that is divisible by 2 and one number  divisible by 4.

We don't have any other primes for which this is true. That is proved simply by the triple 3, 4, 5. All the primes greater than 5 (7, 11, 13, 19, ...) do not divide any of these numbers evenly, so we have a single counter-example that works for an infinite number of cases, a situation a lazy mathematician is always happy to find.