Suppose you have a right triangle and every side has rational length. What can you say about the area? For example, is it possible that such a triangle could have area 1?
It turns out the answer is no, and here’s a proof. Suppose you had a right triangle with sides of length a, b, and c with c being the length of the hypotenuse. And suppose a, b, and c are all rational numbers.
Start with the equation
(a2 – b2)2 = (a2 + b2)2 – 4a2b2
Suppose the area of the triangle is 1. Then ab/2 = 1 and so ab = 2. Use this and the Pythagorean theorem to rewrite the right side of the equation above. Now we have
(a2 – b2)2 = c4 – 16
But this result contradicts a theorem of Fermat: in rational numbers, the difference of two fourth powers cannot be a square.
So a rational right triangle cannot have area 1. Also, it cannot have area r2 for any rational number r. (If it did, you could divide each side by r and have a rational triangle with area 1.)
Now things are about to get a lot deeper.
What values can the area of a rational right triangle take on? Euler called such numbers congruent. Determining easily testable criteria for congruent numbers is an open problem. However, if the Birch and Swinnerton-Dyer conjecture is correct, then an algorithm of Jerrold Tunnell resolves the congruent number problem. (Incidentally, the Clay Mathematics Institute has placed a $1,000,000 bounty on the Birch and Swinnerton-Dyer conjecture.)
What if instead of limiting the problem to rational right triangles we considered all triangles with rational sides? See this paper for such a generalization.