# Curious Identities In Pythagorean Triangles

A problem has been posted in The American Mathematical Monthly (606, 1933), with a solution by Maud Willey (Long Beach, Mississippi) published next year (Vol. 41, No. 5 (May, 1934), p. 330):

Give a scheme for writing down mechanically the sides of an unlimited number of dissimilar right triangles whose sides are integers. After the first set, the values are to be written down, not merely indicated, without any calculations whatever. No addition, subtraction, multiplication, division, involution or evolution, mental or otherwise, is allowed.

One way is to start with the triangle whose sides are 21, 220 and 221.

The problem has been also discussed by W. Sierpinski in Pythagorean Triangles (Dover, 2003).

Solution ### Number Curiosities Give a scheme for writing down mechanically the sides of an unlimited number of dissimilar right triangles whose sides are integers. After the first set, the values are to be written down, not merely indicated, without any calculations whatever. No addition, subtraction, multiplication, division, involution or evolution, mental or otherwise, is allowed.

One way is to start with the triangle whose sides are 21, 220 and 221.

### Solution

The suggested Pythagorean triple $21$, $220$ and $221$ offers a couple clues. First, the even leg is the longer of the two. Second, the long leg is $1$ less than the hypotenuse. let's try to following those clues.

We are thus looking for integers $a$, $b$, $c$ that satisfy $a^{2}+b^{2}=c^{2}$, $b+1=c$, and $a=2n+1$. That's a start.

$(2n+1)^{2} = (b+1)^{2} - b^{2} = 2b+1$, implying $b=2n^{2}+2n$. Since $b+1=c$, $c= 2n^{2}+2n+1.$  What remains is to choose $n$ to be a power of $10$, $n=10^{k}.$  The given triple is obtained with $k=1.$  For $k\gt 1,$  the triples are obtained by inserting zeros - an operation permitted by the conditions of the problem:

\begin{align} a & & b & & c \\ 21 & & 220 & & 221 \\ 201 & & 20200 & & 20201 \\ 2001 & & 2002000 & & 2002001 \\ 20001 & & 200020000 & & 200020001 \\ \cdots \end{align}

Adopting a subscript to indicate the number of repeated digits, the general rule could be written as

$(20_{t}1)^{2} + (20_{t}20_{t+1})^{2} = (20_{t}20_{t}1)^{2}$.

Taking $n=2\cdot 10^{k}$ leads to

$(40_{t}1)^{2} + (80_{t}40_{t+1})^{2} = (80_{t}40_{t}1)^{2}$.

Observe that due to the condition $c=b+1$ all the triples obtained in this manner are primitive.

There is another family of triangles:

$(60_{t}9)^{2} + (20_{t}60_{t+1})^{2} = (20_{t}60_{t}9)^{2}$.

But this is a completely different story. 