Inspired by Fibonacci

Number Theory Level 5

Does there exist a right triangle with three rational sides whose area is 5? If so, find the least common denominator of the sides; if not, enter 666.

Bonus question : How many such triangles are there?

The answer is 6.

3 solutions

Michael Mendrin
Nov 25, 2015

Well, that was an easy one. But it's interesting problem! "Inspired by Fibonacci" (?) is what makes it interesting.

It's about finding integers $k, m, n, x$ , where $GCD(m,n)=1$ and $a=\dfrac{k}{x}(2mn)$ , $b=\dfrac { k }{ x } \left( { m }^{ 2 }-{ n }^{ 2 } \right)$ such that

$\dfrac { {k}^{2} }{ { x }^{ 2 } } mn(m-n)(m+n)=5$

$k,x,m,n=1,6,5,4$ is one solution, so that $a=\dfrac {40}{6}$ and $b=\dfrac{9}{6}$

Yes, that's the answer Fibonacci gave when this problem was submitted to him by the court of Kaiser Friedrich II in 1225.

One follow-up question: How do we know that 6 is the least common denominator?

In the case of a small number like 5, we can get away with a straightforward search like this... but how would we handle a number like 2015? ;)

Bonus question: How many such triangles are there?

Otto Bretscher - 5 years, 6 months ago

"In the case of a small number like 5" is the key to being able to solve this one! Somehow, a lot of factors just has to vanish on the left side, and we already know either $m$ or $n$ has to have $5$ as a factor. That greatly limits our search.

How would we handle $2015$ ? I don't know, so, come on, please post it. I'm gonna make it a threat---if you don't post it, I will (as soon as I figure out the answer myself). I don't think we should be afraid to post really hard problems on Brilliant.org.

"How many such triangles are there?"----okay, if that is actually a finite number, that sounds like a really tough Brilliant.org problem!

Edit: Regarding posting a problem in the case of $2015$ , the answer could be so large that it'd be impractical to make it a Brilliant.org problem. I can see that happening. I don't know the number yet, but I'm already starting to get that hunch. For "right triangles with rational sides and integer areas", as the integer area increases, the rational fractions seem to grow very quickly. I'm starting to think that a Note is the better place to treat this general problem.

Michael Mendrin - 5 years, 6 months ago

For the benefit of the other readers, we should make it clear why 2,3,4, or 5 aren't possible common denominators, since I asked for the least common denominator.

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – We can pretty much immediately decide that $k=1$ and that $m=5$ , so that we're left with this to solve for integer $n,x$

$n(5-n)(5+n)={x}^{2}$

A quick scan for integers $n>0$ gets us $n=4, x=6$ as the smallest possible $x$

For larger areas, the problem becomes massively more difficult. I can see it involves elliptic equations, like the kind in this problem Playing With An Elliptic

Michael Mendrin - 5 years, 6 months ago

@Michael Mendrin – Yes, a scan certainly works. Alternatively, we could point out that $(xa,xb,xc)$ will define a Pythagorean triangle with area $5x^2$ . Since the area of a Pythagorean triangle is always divisible by 6, we can conclude that $x$ must be divisible by 6 as well.

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Otto, you know, I went off on a wrong tangent thinking this was about integer sides, and I hadn't been thinking too deeply about the far more general problem of right triangles with rational sides. Now that I'm looking into this, well, let me say this deserves or at least requires a full Note by itself. This is a very complicated subject! I think I understand now why you haven't been willing to post a problem in the case of $2015$ . The solution probably wouldn't fit in here.

Michael Mendrin - 5 years, 6 months ago

@Michael Mendrin – There is a pretty simple criteria for the existence of a rational right triangle of area $n$ , stated in Tunnell's theorem: You just have to count the integer points on two ellipsoids. But there is a lot of (beautiful!) theory behind it, and one part of Tunnell's theorem assumes the (unproven) BSC conjecture.

I believe the general problem "Find an algorithm to decide whether a given $n$ is the area of a rational right triangle" is still open, 800 years after Fibonacci first posed it (and it appears that Arab mathematicians have been thinking about this issue even earlier).

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Thanksgiving is coming up and, well, I'm kind of tied up with that now. I'll try to get back to this later tonight. Yes, it's a very interesting subject, and I have some experience with elliptic curves, so I'll have a closer look at this.

Off hand, in just fooling around with this problem, I think there are just 2 possible solutions. But I could be wrong.

Michael Mendrin - 5 years, 6 months ago

@Michael Mendrin – Take your time, and enjoy your Thanksgiving!

I suspect that there are infinitely many rational right triangles with area 5 (or 2015), based on an argument involving elliptic curves, but I will have to examine the details of my proof.

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – I'm calling it a day, and you enjoy your Thanksgiving as well, Otto.

Michael Mendrin - 5 years, 6 months ago

@Michael Mendrin – Ah, I see that the conversation has continued here.

I studied congruent numbers as an undergrad, and I recall the following results:

Accounting for sign changes, there is a 1-1 correspondendence between rational right triangles with area $n$ , and rational solutions to $y^2 = x^3 - n^2x , y \neq 0$ .
By the theory of elliptic curves, if there is one (non-trivial) solution, then there are infinitely many (just add elliptically).

(The trivial solutions are $(x,y) = (0,0), (n,0) (-n, 0)$ . )

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – I'm glad to see that I was on the right track. In "More fun in 2016, Part 5", I posted a very similar solution. Thanks for confirming that this approach works!

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Yup, that's the approach (assuming that one understands the machinery behind it).

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – Look above in my replies to Otto about a problem I've posted a while back about elliptic curves. So, the 2nd bullet point is noted.

What was your major in undergrad college?

Michael Mendrin - 5 years, 6 months ago

@Michael Mendrin – Mathematics and economics. (What else would you expect?)

I was investigating congruent numbers for a book that my professor was writing up, explaining to high school students how they can be working on such questions, which quickly lead into areas of research.

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – Tell us about that book, please!

Otto Bretscher - 5 years, 6 months ago

@Michael Mendrin – Yes, exactly, elliptic curves... that's how I got interested in this problem. There is an obvious connection to the rational points on $y^2=x^3-n^2x$ , where $n$ could be 5 (easy) , 2015 (hard) or 2016 (intermediate). Here is an idea for a new problem: Find a nontrivial (meaning $y$ is nonzero) integer point on the curve $y^2=x^3-2016^2x$ .

Otto Bretscher - 5 years, 6 months ago

Worranat Pakornrat
Nov 26, 2015

Let a and b be the lengths of the shorter sides of the right triangle and c be the length of the hypotenuse side. Since the area is 5, then ab = 10.

Then b = $\frac{10}{a}$ .

Suppose a = $\frac{x}{y}$ for some integers x, y as a is rational.

Then $c^{2}$ = $a^{2}$ + $b^{2}$ = $a^{2}$ + $\frac{100}{(a^{2})}$ = $\frac{a^4+100}{(a^{2})}$

Taking a square of c, the denominator will equal to a, so $a^4$ +100 must be perfect square.

Then $a^4$ +100 = $(\frac{x}{y})^4$ +100 = $\frac{x^4+100y^4}{(y^{4})}$

Taking a square of this, it will leave $y^2$ as a denominator, so $x^4$ +100 $y^4$ is still a perfect square.

Now suppose $d^2$ = $x^4$ +100 $y^4$ = $(x^2)^2$ + $(10y^2)^2$ for some integer d.

That means the integers d, $x^2$ , & $10y^2$ form a special Pythagorean triple, where one side is a perfect square long while the other is a multiple of 10 and a perfect square number.

Going for the right order, we will find that the triple (9, 40, 41) fits the criteria, thereby yielding x = 3 and y = 2.

Thus, a = $\frac{3}{2}$ , and b = $\frac{20}{3}$ , and c = $\frac{41}{6}$ . Hence, the least common denominator of these fractions is 6.

Toby M
Apr 22, 2019

Let the legs of the triangle be $a$ and $b$ , and the hypotenuse be $c$ . Then $ab$ = 10. If we can find a Pythagorean triplet with the product of the legs as $10k^2$ , with $k$ being an integer.then we can scale the legs down by $k$ and still have a Pythagorean triple. We only have to search for primitive Pythagorean triples since if the product is not in the form $10k^2$ , scaling the sides up by an integer will not change this.

Searching through $(3,4,5)$ , $(5, 12, 13)$ , and $(7, 24, 25)$ , we find that the product of the legs of $9, 40, 41$ is $360$ , so we have to scale the legs down by $\sqrt{\frac{360}{10}} = 6$ . Therefore, $a = \frac{9}{6}$ , $b = \frac{40}{6}$ , and the least common denominator is $6$ .

Inspired by Fibonacci

The answer is 6.

3 solutions

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