Maximum ratio of the shorter to the longer leg in a Pythagorean triangle

Question

Problem

I was reading the question Why is the smallest Pythagorean triple $(x,y,z)=(3,4,5)$ not close (in ratio $x/y$) to any other small triple?. I came up with this question:

Let $(a, b, c)$ be a Pythagorean tuple where $a < b < c$. What is the maximum possible value of $a / b$?

(A triple $(a, b, c)$ is a Pythagorean triple if $a, b, c \in \mathbb{Z}^+$ and $a^2 + b^2 = c^2$)

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My attempt

I am aware of Euclid's formula$^1$ $$ a = 2mn, \qquad b = m^2 - n^2, \qquad c = m^2 + n^2. \qquad (m,n \in \mathbb{Z}^+, m > n)$$ Therefore, $$ \frac{a}{b} = \frac{2mn}{m^2 - n^2}. $$ Now, how can I maximize it? I can't see a lot of connection between the numerator and the denominator.

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$^1$ It turns out that this step is wrong — it fails to consider the case where $2mn > m^2 - n^2$.

Answer 1

There are infinitely many Pythagorean triples $(a, b, c)$ with $b=a+1$, see for example

• Wikipedia: Pythagorean triple:

  Choosing m and n from certain integer sequences gives interesting results. For example, if m and n are consecutive Pell numbers, a and b will differ by 1

• The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$

An infinite set of positive integers is necessarily unbounded, therefore $$ \frac{a}{b} = 1 - \frac{1}{a+1} $$ can be arbitrarily close to one for these triples.

Answer 2

We will show that the maximum ratio of the short leg to the long leg is $3/4$ by contrasting it with the ratio of the long leg to the short leg.

For Pythagorean triples (if we insist that side-A is odd) it is not always the case that $A<B<C$. For example $(15,8,17)\quad (35,12,27)\quad (63,16,65)...\quad$

If in Euclid's formula $ (A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2)$ we let $(m,k)$ be $(2n-1+k,k)$ and let $k=1$ we arrive at a formula that makes it easy to see how $A/B$ can be maximized.

$$A=4n^2-1\quad B=4n\quad C=4n^2+1\implies R=\frac{A}{B}=f(n)=\frac{4n^2-1}{4n}$$ Here are examples of $F(n)\rightarrow (A,B)$ represented as $n(A,B)$ $$1(3,4)\quad 2(15,8)\quad 3(35,12)\quad 4(63,16)\quad ...\quad 25(2499,100):(R=24.99)$$

We can see by this progression that the ratio $RnA/B$ can grow indefinitely with $n$.

$$\lim_{n\rightarrow\infty} \bigg(\frac{4n^2-1}{4n}\bigg)=\infty$$

As an alternative, if we let $(m,k)$ be $(2n-1+k,k)$ and $n=1$ we find that $B/A$ grows faster.

$$A=2k+1\quad B=2 k^2 + 2 k\quad C=2 k^2 + 2 k + 1 \implies R=\frac{B}{A}=f(k)= \frac{2 k^2 + 2 k}{2k+1}$$ where $$1(3,4)\quad 2(5,12)\quad 3(7.4)\quad 4(9,40)\quad ...\quad 25(51,1300) :(R\approx 25.49)$$

$$\lim_{k\rightarrow\infty} \bigg(\frac{2 k^2 + 2 k}{2k+1}\bigg)=\infty$$

All other combinations of $(n,k)$ are slower to grow but all Ratios approach infinity. In any case, the first triple in any series has the largest ratio of short leg to long leg. $$\therefore R_{max}=\frac{3}{4}$$