Why Must One Leg of a Primitive Pythagorean Triple Be a Multiple of 3?

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In summary, it has been proven that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3. This is because if both a and b are not multiples of 3, then c must be a multiple of 3, and if one of a or b is a multiple of 3, then c cannot be a multiple of 3. This is due to the fact that the sum of the squares of two integers cannot be a multiple of 3 if both integers are not already multiples of 3.
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alexmahone
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Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3.

My attempt:

Let a and b be relatively prime positive integers.

If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$

This is impossible as the only quadratic residues modulo 3 are 0 and 1.

So far, so good.

If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$

This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.
 
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Alexmahone said:
Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3.

My attempt:

Let a and b be relatively prime positive integers.

If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$

This is impossible as the only quadratic residues modulo 3 are 0 and 1.

So far, so good.

If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$

This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.

you have established that
If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$
you are right that $c^2 = 1 \pmod{3}$ does not mean that $c^2$ is perfect square but in the above you have shown that for a Pythagorean triplet above condition must be true. it is one way and not both ways
 

FAQ: Why Must One Leg of a Primitive Pythagorean Triple Be a Multiple of 3?

What is a Primitive Pythagorean triple?

A Primitive Pythagorean triple is a set of three positive integers (a, b, c) that satisfy the Pythagorean theorem, a^2 + b^2 = c^2, and where a, b, and c have no common factors.

How many Primitive Pythagorean triples are there?

There are infinitely many Primitive Pythagorean triples. However, they can be generated using the formula a = m^2 - n^2, b = 2mn, c = m^2 + n^2, where m and n are any positive integers with m > n and m and n have no common factors.

What is the significance of Primitive Pythagorean triples?

Primitive Pythagorean triples have been studied for centuries and have many applications in mathematics, especially in number theory and geometry. They also have connections to other fields such as cryptography and physics.

How can we determine if a set of three numbers is a Primitive Pythagorean triple?

To determine if a set of three numbers (a, b, c) is a Primitive Pythagorean triple, we can use the following criteria:

  • a, b, and c must be positive integers
  • a^2 + b^2 = c^2
  • a, b, and c must have no common factors (they must be relatively prime)

Are there any special properties of Primitive Pythagorean triples?

Yes, there are several special properties of Primitive Pythagorean triples. One example is that the sum of any two numbers in the triple is always odd. Another is that the sum of the squares of the two smaller numbers is equal to the square of the largest number.

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