3
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your proof is not correct....side lengths are rational....not integer...so you cannot assume $x=2,y=1$...there are infinite number of rational cases possible.... $\endgroup$ – tatan Oct 27 '16 at 16:52
  • $\begingroup$ Sorry. I misunderstood $\endgroup$ – Marvel Maharrnab Oct 27 '16 at 21:41
  • $\begingroup$ Tatan now see the new proof is it ok $\endgroup$ – Marvel Maharrnab Oct 27 '16 at 22:02
  • $\begingroup$ Yeah ok...but you should try to improve your answer by adding some more data on Fermat's Theorem...I will say your answer is correct but not well written... $\endgroup$ – tatan Oct 28 '16 at 3:28
  • $\begingroup$ Brother see the link given by gerry myerson $\endgroup$ – Marvel Maharrnab Oct 28 '16 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.