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.

17 thoughts on “Rational right triangles

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

    Should be,

    (a2 – b2)2 = (a2 + b2)2 – 4a2b2

  4. 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)

  5. 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.

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

  7. The formula is a rearrangement of the classic formula for Pythagorean triples (see https://en.wikipedia.org/wiki/Pythagorean_triple). If a, b, and c are integer sides of a right triangle, then a = m^2 – n^2, b = 2*m*n, c = m^2 + n^2 for some integers m and n (i.e., (m^2 – n^2)^2 + 4*m^2^n^2 = (m^2 + n^2)^2).

    I tried playing with this to see if one could adapt this proof to get a more motivated proof, but I couldn’t get it to work.

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