Number of integer triangles. 
Number of integer isosceles or equilateral triangle none of whose
  sides exceed $2c$ is?

I substituted $c =3$ checked, and got $27= 3 \times (3^2)$ i.e. $3c^2$ triplets. It gives the right answer($3c^2$) but how do I write a proper formal proof of this?
I tried it this way: We have three vacant places for three integers. First place can be filled in $2c$ ways, 2nd one in $2c$ ways, 3rd one in $2c$ way but it involves over-counting, is it possible to eliminate the extras? 

Seems like a recursive approach would work well.
The number of isoceles triangles with the two equal legs equal to $2c$, and the other shorter, is of course $2c{-}1$.
The number of isoceles triangles with the two equal legs equal to $2c{-}1$, and the other shorter, is $2c{-}2$.
The isosceles triangles with "base" of $2c$ and equal legs shorter is $c{-}1$, since the equal legs must be more than $c$ in length, and likewise with a base of $2c-1$ the count is also $c{-}1$ (since equal legs of $c$ will now qualify).
Adding the two equilateral triangles of sides $2c$ and $2c-1$, we get $3c{-}1$ triangles on the increment between $c{-}1$ and $c$.
So our cumulative value $t(c) = t(c-1) + 6c-3$, with $t(0)=0$ of course. Then the values $3,12,27,48,75,\ldots$ can be assessed to give $t(c) = 3c^2$.
