A number theory problem by Santarpan Roy

To form the largest sided right ∆, 12 needs to be the smallest side.

It is achieved when the other two sides are as close to each other as possible.

Let's start with difference between other two sides (a and b) =1. We know difference between their squares is then 2n+1 where n is the smallest side among a and b. But 2n+1=144 gives a non integral n.

So we move to difference between a and b=2. So $2n+1+2(n+1)+1=144.$ This gives $n=35$ and $n+2=37.$

So maximum perimeter is $12+35+37=\boxed{84}$ :)

let's define the sides x ,12 and the hypotenuse 12. accroding to P theorom:x^2+144=y^2, (y-x)×(y+x)=144 let's assume: y+x=a, y-x=b solving for y we get: y=(a+b)/2. so we know that a and b are either even or non even integers. 144=12^2= (3^2)×2^4 so it's impossible to find 2 non even Factors which leaves us knowing that a and b are both even Integers since we want Max parimeter we need to find max y,x so "a" should be maximal and "b" minimal and the choice a=72 b=2 setisfy the conditions leaving us with x=35 and y=37. and therefore the answer is 35+37+12=84.