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This is the second of the two articles on right-angled triangles whose edge lengths are whole numbers. We suppose that the lengths of the two sides of a right-angled triangles are $a$ and $b$, and that the hypotenuse has length $c$ so that, by Pythagoras' Theorem,$$a^2 + b^2 = c^2$$.
In the first article we discussed the possibility of enlarging or shrinking a right-angled triangle to get another right-angled triangle whose sides also have lengths that are whole numbers, and we claimed there that apart from a possible scaling of the triangle, every such right-angled triangle has edge lengths of the form
| $a=2pq \; \; \;$ |
$b=p^2-q^2 \; \; \;$ |
$c=p^2+q^2$
|
For suitable whole numbers $p$ and $q$, where $p> q$. This article contains a proof of this fact.
First, we must understand a few ideas about factors and prime numbers. A factor of a whole number $n$ is a whole number $f$ that divides into $n$ exactly (without remainder). Of course, $1$ and $n$ are always factors of $n$, and we say that $n$ is a prime number if $1$ and $n$ are the only factors of $n$. We do not regard $1$ as a prime number (even though its only factor is 1), and you
should now check that the first few prime numbers are $2$, $3$,$5$, $7$, $11$, $13$, $17$.
What are the next three prime numbers?
If a number $n$ is not a prime number then we must be able to write it as a product of two numbers $u$ and $v$; that is $n = u \times v$, and we normally write this as $n = u v$. If $u$ (or $v$) is not a prime number, then we can write $u$ (or $v$) as a product and so write $n$ as a product of three numbers. We can continue in this way until every number that we are left with is a prime
number, and this shows that every number $n$ is a product of prime numbers.
For example, suppose that we start with $255$. Then $255 = 5 \times 51$. Now $5$ is a prime, but $51$ is not. Next, $51 = 3 \times 17$ and both $3$ and $17$ are primes; thus $255 = 3 \times 5 \times 17$ and we say that $3$, $5$ and $17$ are the prime factors of $255$. Of course, some prime factors may be repeated; for example, $75 = 3 \times 5 \times 5$, and $315 = 3 \times 105 = 3 \times 5
\times 21 = 3 \times 3 \times 5 \times 7$.
If we know that $f$ is a prime factor of a product $u v$, then (writing $u$ and $v$ as a product of prime numbers) we see that $f$ must occur as one of the prime factors of $u$ or of $v$ (or of both), so that $f$ must be a factor of $u$ or of $v$.
We repeat :
(I) if $f$ is a prime factor of $u v$ then $f$ must be either a prime factor of $u$ or a prime factor of $v$.
Note that this result is NOT true of every factor; our claim applies only to prime factors. For example, $6$ is a factor of $4 \times 9$, but it is not a factor of $4$ or of $9$; of course, each prime factor of $6$ is a factor of either $4$ or $9$.
A number $n$ is a square number if $n = m^2$ for some whole number $m$. It is easy to see (by writing $n$ as the product of its prime factors) that a whole number $n$ is a square number if every one of its prime factors occurs an even number of times. For example:
$5 \times 5 \times 7 \times 7$ is a square number but $3 \times 5 \times 7 \times 7$ is not.
Here are two useful facts about factors and square numbers.
Suppose that $f$ is a prime factor of the square number $n=m^2$. Then
(II) $f$ is a prime factor of $m$, and
(III) $f$ is a prime factor of $n$.
The statement (II) is just the statement (I) with $u = v = m$.
From (II) we see that $f$ is a prime factor of $m$, and this means that $f$ is a factor of $m^2 = n$ which is (III).
We now return to the problem of showing that every triple of whole numbers $a$, $b$, $c$ with $a^2+ b^2 = c^2$ can be expressed in the form (1).
To show this, we start with any Pythagorean triple and first reduce it as much as possible to end with a triple $a$, $b$, $c$ which cannot be reduced any more. This means that there is no whole number (except 1) which is a factor of each of $a$, $b$, $c$ (for otherwise, we could reduce the triangle still further). We shall show now that these $a$, $b$, $c$ can be written in the form (1) for
some suitable $p$ and $q$. There are, of course, three possibilities that can arise, namely:
(i) $ a$ and $b$ are both even;
(ii) $a$ and $b$ are both odd;
(iii) one of $a$ and $b$ is even and the other is odd.
In fact, neither (i) nor (ii) can happen so let us see why.
First, (i) cannot happen because if $a$ and $b$ are both even, then $c^2 (=a^2 + b^2)$ is even, so that $c$ is also even. But then $a$, $b$ and $c$ each have a factor 2 and we could have reduced the triple $a$, $b$, $c$ still further.
So $a$ and $b$ cannot both be even.
To see that (ii) cannot happen suppose that $a$ and $b$ are both odd. Then the remainder when dividing $a^2$ (and also $b^2$ ) by $4$ is $1$, so that the remainder when dividing $c^2 (= a^2 + b^2 )$ by 4 is 2.
However, if $c$ is even, $c^2$ is a mulitple of 4, so this remainder must be 0, while if $c$ is odd the remainder must be 1.
We now know that (iii) must hold and we shall take $a$ to be the even number and $b$ to be the odd number.
As $c^2 = a^2 + b^2$ , we see that $c$ must be odd.
As $b$ and $c$ are odd, their sum must be even and their difference must also be even. So we can find whole numbers $r$ and $s$ such that $2r = c + b$ and $2s = c - b$, and this means that
\begin{eqnarray} b = r - s,\\ c = r + s \\ \mbox{ and } \mbox{ } a^2 = c^2 - b ^2 = (c + b)(c - b) = 4rs \mbox{. . . . . . . (2)} \end{eqnarray}
This is similar to (1) but not quite the same. Also, not every choice of $r$ and $s$ here gives us a triple of whole numbers. For example, if we take $r=2$ and $s=1$ we get $a^2 = 8$ so that in this case $a$ is not a whole number. We are going to show that $r$ and $s$ in (2) must be square numbers. Then we can write $r=p^2$ and $s=q^2$, say, and then (2) gives
\begin{eqnarray} a = 2p q \\ b = p - q \\ c = p + q \mbox{ which is (1).}\end{eqnarray} The hardest part of this article is to show that $r$ and $s$ are square numbers, and this depends on understanding the ideas about factors and prime numbers.
First, we must show that $r$ and $s$ cannot have any common factors.
To see this, suppose that $r$ and $s$ have a common prime factor $f$. Then $r - s (= b)$ and $r + s (= c)$ both have $f$ as a factor, so that $f$ is a factor of $c^2 - b^2$, which is $a^2$ . As $f$ is a prime factor of $a^2$, then $f$ is a factor of $a$.
This would mean that $f$ is a factor of each of $a$, $b$, $c$ and this cannot be so else again we would have reduced the triple $a$, $b$, $c$ still further at the outset.
We have now shown that $r$ and $s$ have no common factors.
Now take any prime factor $f$ of $r$. First, as $r$ and $s$ have no common factors, we see that $f$ is not a factor of $s$. Next, as $f$ is a prime factor of $r$, it is also a prime factor of $r s$, and hence a prime factor of the whole number $\left( \frac{a}{2}\right)^2$ (because $\left(\frac{a}{2}\right)^2 = r s$).
As $f$ is a prime factor of $\left(\frac{a}{2}\right)^2$, it follows from (III) that $f^2$ is a factor of $\left(\frac{a}{2}\right)^2 = r s$, and because $f$ is not a factor of $s$, we now see that $f^2$ is a factor of $r$.
We have just seen that if $f$ is any prime factor of $r$, then $f^2$ is a factor of $r$, and this means that $r$ is a square number. The same argument is true for $s$ as well as $r$, so that as $r$ and $s$ are square numbers.
Because $r$ and $s$ are square numbers we can now write $r=p^2$ and $s=q^2$ for some $p$ and $q$ and, at last, we have shown that $a = 2p q$, $b = p^2 - q^2$ and $c = p^2 + q^2$ which is (1).