Rational right triangles

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

(a2b2)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

(a2b2)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.

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13 comments on “Rational right triangles
  1. marinamedi says:

    You write:

    Start with the equation

    (a2 – b2)2 = (a2 + b2) – 4a2b2

    Can you explain me where does that expression?

    I think this article would be interesting for my maths class in university

    Greats from a spanish twitter follower. Thanks

  2. John says:

    So, why start with (a2b2)2 = (a2 + b2)2 – 4a2b2? I admit that doesn’t seem very motivated, but it leads to a short proof. I don’t know of a more natural proof.

    Update: I may have misunderstood your question. I realize now there was a typo in the initial equation. I assume you were asking why it was correct — it wasn’t! — not where it came from.

  3. Neeraj says:

    John,
    I believe there is a typing mistake.
    (a2 – b2)2 = (a2 + b2) – 4a2b2

    Should be,

    (a2 – b2)2 = (a2 + b2)2 – 4a2b2

  4. There is a superscript 2 missing from the right-hand side of the first equation, which is probably why it is puzzling the previous commenters.

    I would be interested in an expert mathematician’s view on whether this result by me:
    http://gpj.connectfree.co.uk/gpjk.htm#(5)
    has any relevance to this subject.

  5. marinamedi says:

    still don’t understand. Can the author or someone answer me?
    thanks a lot

  6. A. Webb says:

    Marinamedi, the first equation is just rearrangement using algebra.

    To make this easier to see, let s = a^2 and t = b^2. All this equation is saying is that

    (s – t)^2 = s^2 – 2st + t^2 (expand)
    = s^2 + 2st – 4st + t^2 (add and subtract 2st)
    = (s^2 + 2st + t^2) – 4st (rearrange)
    = (s+t)^2 – 4st (factor first expression)

  7. Mitch says:

    marinamedi, this is an extremely unmotivated proof. It proves the theorem, but without at first looking like it had anything to do with the theorem. The first equation has nothing to do with a triangle it is just an identity of polynomials (expand both sides algebraically). Then an algebraic fact about a particular a and b (of a right triangle) is used to get the second equation. Then Fermat’s -algebraic theorem applies (by contradiction). How this proof was found (I have no idea!) is not relevant, the proof still works, a and b can’t be 1. So in some sense, your question really is ‘how was this particular proof found?. and that I have no idea about.

  8. marinamedi says:

    Thanks Mitch and A.Webb. You had helped me a lot. ¡Gracias!

  9. Aleksandr Sayko says:

    What about triangle with side of length 3, 4 and 5. It seems to be a contraexample.

  10. John says:

    Aleksandr: Such a triangle has area 6.

  11. Mitch says:

    Aleksandr: intuitively that seems right but to go from area 6 to area 1, you have to divide the -lengths- by sqrt(6).

  12. Aleksandr Sayko says:

    I see. It’s like to be an interesting fact. And what about not right triangles?

  13. John says:

    See the last line of the post for a link to a paper discussing general triangles.

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