More fun in 2016, Part 5

We can use Euclid's formula for Pythagorean triples, $a=2mn$ and $b=m^2-n^2$ for the legs and $c=m^2+n^2$ for the hypotenuse. The only integer solution of $mn(m^2-n^2)=2016$ is $m=9, n=7$ , with $c=\boxed{130}$ .

An integer which is the area of a Pythagorean triangle with rational sides is called congruent . The bonus question asks if $2015$ is congruent. The Congruent Number Problem, determining whether a particular integer is congruent or not, is one of the big unsolved problems in Number Theory.

Assuming the Birch-Swinnerton Dyer Conjecture, it has been proved that all square-free integers congruent to any of $5,6,7$ modulo $8$ are congruent. This would make it highly likely that $2015$ is congruent. The number $2015$ is listed as congruent by Noe in a link that can be followed from OEIS, A003273.

Although I do not have an explicit Pythagorean triad to demonstrate it, I am led to believe that $2015$ is congruent.

guys, one thing you should know is as one side gets bigger, the other gets smaller, but notice that $huge^2+small^2\neq some^2$ .we will start from the middle hence. first, let adj=a,opp=b, hyp=c. then $\dfrac{ab}{2}=2016\Longrightarrow ab=4032$ so lets check from middle: $\begin{array}{c}&\quad\quad\quad\quad\quad\quad a,b=(64,63)\Longrightarrow c\notin \mathbb{N}\\ (a,b) = (192,21)\Longrightarrow c\notin\mathbb{N}\quad (a,b)=(32,126)\Longrightarrow c=130\end{array}$ it is obvious that further down either path will not result in any more rational 'c's. so we conclude $\boxed{130}$ is the only and the biggest 'c'(hypotenuse).

Using the classic generator for Pythagorean triples, this would have to be true

$\dfrac { 1 }{ 2 } (2kmn)k({ m }^{ 2 }-{ n }^{ 2 })=2015$

so that we'd quickly end up with this, since $2015$ is square free

$mn(m-n)(m+n)=5\cdot 13 \cdot 31$

which won't work out