Find a triangle whose three sidelengths and area and perimeter are all square numbers.
Taking $a^2,b^2,c^2$ be the sidelengths of the triangle,then we need to find positive integer solutions to this equation: $$a^2+b^2+c^2=x^2,\\ x^2(x^2-2a^2)(x^2-2b^2)(x^2-2c^2)=y^4.$$
Let $x=1$ then we need to find positive rational solutions to this equation: $$a^2+b^2+c^2=1,\\ (1^2-2a^2)(1^2-2b^2)(1^2-2c^2)=y^4.$$