Right angled triangles with integer areas. I've recently been working on the following problem:

Is there a right angled triangle with rational side lengths and an area of $1$?

I was told I wouldn't be able to solve it, but this was simply an exercise to see how I go about solving problems.
I indeed was not able to solve, but was able to reduce it to a problem with an elliptic curve that I suspect relates to Fermat's last Theorem. However, I suspect there is a more subtle and simpler approach to the problem. So what exactly is the answer?
 A: I will give a hint.Let one side of the triangle be a/b.As i had mentioned above to get the area as one the length of 2nd side must be 2*b/a.We get the hypotenuse as  sqrt(4b^4+a^4)/ab.Since we take the a,b values as integers we get the denominator a integer value.now solve the numerator to get an integer value which give you yhe required answer
A: Let the right triangle have sides of length $a$, $b$, and $c$ (where $c$ is the length of the hypotenuse) and suppose $a$, $b$, and $c$ are all rational numbers.
By previous knowledge we have the equation $(a^2 – b^2)^2 = (a^2 + b^2)^2 – 4a^2b^2$.
We also know the area of the triangle is $1$, so therefore $\frac {ab}{2} = 1$. Multiplying both sides by $2$ gives $ab = 2$.
Using the Pythagorean theorem we have $(a^2 – b^2)^2 = c^4 – 16$. However, by previous knowledge we know a theorem of Fermat which states that the difference of two rational numbers raised to a power of $4$ cannot be a square of a rational number. Therefore, we have a contradiction and the proof is complete.
Special thanks to @tatan @lulu @GerryMyerson for helping in the proof.
