Mr. Pythagoras!

Number Theory Level 3

The area of the triangle formed by $pythagorean$ $triplet$ with integer sides is always divisible by 3 .

$True$ or $False$ ?

False

Sometimes

True

Not

enough

information

1 solution

Md Zuhair
Oct 21, 2017

I think using the primitive pythagorean triplets( $2mn,m^2-n^2,m^2+n^2$ )we can easily manage that. Note that m and n are relatively prime.

Yes, in general if the legs have length $2kmn$ and $k(m^{2} - n^{2})$ then the area of the triangle is $k^{2}mn(m^{2} - n^{2}) = k^{2}mn(m - n)(m + n)$ .

If either $m$ or $n$ is divisible by $3$ then we are done. If not, then if $m,n$ are both equivalent to either $1$ or $2$ mod $3$ then $m - n$ will be divisible by $3$ , and if one is equivalent to $1$ mod $3$ and the other to $2$ mod $3$ then $m + n$ will be divisible by $3$ .

Brian Charlesworth - 3 years, 7 months ago

Mr. Pythagoras!

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