## Pythagorean Triplet Product

#### Problem

A Pythagorean triplet, $(a, b, c)$, is defined as a set of positive integers for which $a^2 + b^2 = c^2$.

Prove that for every triplet $abc$ is a multiple of sixty.

#### Solution

If we can show that it is true for all primitive cases, then it must be true for all non-primitive cases (multiples of primitive triplets).

We shall use the fact that for $m \gt n$, all primitive triplets can be generated with the following identities (see Every Primitive Triplet):

\begin{align}a &= m^2 - n^2\\b &= 2mn\\c &= m^2 + n^2\end{align}

We shall complete this proof by showing that $abc$ must contain a factor of 4, 3, and 5.

• Factor of 4:
If either $m$ or $n$ are even then $b = 2mn$ will be a multiple of 4. If both $m$ and $n$ are odd, then $m^2 + n^2$ will be even, and $bc$ will be a multiple of 4. Hence there will always be a factor of 4 present.

• Factor of 3:
If either $m$ or $n$ are multiples of 3 then $b = 2mn$ will be a multiple of 3. Next we note that for each $m,n \equiv 1, 2 \mod 3$, that $m^2$ can only be congruent with 1. Therefore $a = m^2 - n^2 \equiv 0 \mod 3$. Hence there will always be a factor of 3 present.

• Factor of 5:
If either $m$ or $n$ are multiples of 5 then $b = 2mn$ will be a multiple of 5. Next we note that for each $m, n \equiv 1, 2, 3, 4 \mod 5$, that $m^2$ can only be congruent with 1 or 4. Therefore, either $a = m^2 - n^2 \equiv 0 \mod 5$ or $c = m^2 + n^2 \equiv 1 + 4 \equiv 0 \mod 5$. Hence there will always be a factor of 5 present.

Thus we demonstrate that $abc \equiv 0 \mod 60$ for all Pythagorean triplets.

Problem ID: 301 (02 Jan 2007)     Difficulty: 3 Star

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