\[a^2 + b^2 = c^2.\]
If $a, b, c$ are coprime we call the triple a Primitive Pythagorean Triple, or PPT for short. Well known examples are the triples $(3, 4, 5)$ and $(5, 12, 13)$. There is a huge literature about PPTs: how we can generate them in a systematic way, what kinds of number theoretic properties they possess, and so on. Here is a typically striking result about PPTs:
- If $(a, b, c)$ is a PPT, then $a b c$ is divisible by $60$.
Let us now tweak the problem a bit, and ask about triples of positive integers that almost satisfy the condition required of a Pythagorean triple, missing it by the smallest possible margin (namely, by $1$). Thus, we want triples $(a, b, c)$ of positive integers such that
\[a^2 + b^2 = c^2 \pm 1.\]
If the positive sign holds we shall call $(a, b, c)$ an APT - short for "almost Pythagorean triple". And if the negative sign holds we shall call $(a, b, c)$ a NPT - short for "near Pythagorean triple".
But finding APTs and NPTs turns out to be trickier than finding PPTs. The difficulty is caused by the fact that the defining equations are not homogeneous. In the case of PPTs the defining equation is homogeneous of degree $2$; if we divide the equation $a^2 + b^2 = c^2$ by $c^2$, and write $x = a/c$, $y = b/c$, we get the equation $x^2 + y^2 = 1$, whose form immediately brings to mind the equation of a unit circle in the Cartesian plane. This observation if pursued yields a parametric solution to the Pythagorean equation. It is just this approach that has been followed in the articles listed above.
But this approach fails in the case of APTs and NPTs; the non-homogeneity of the equation $x^2 + y^2 = z^2 \pm 1$ negates all such attempts. So for the moment let me not say how we can go about finding them, but simply exhibit a few such triples - found simply by "brute force search" using a computer.
Here are some APTs. For simplicity I have taken $a \le b \le 30$.
a b c
_________
4 7 8
5 5 7
6 17 18
7 11 13
8 9 12
9 19 21
10 15 18
11 13 17
11 29 31
13 19 23
14 17 22
15 26 30
16 23 28
17 21 27
19 27 33
20 25 32
23 29 37
29 29 41
(Comment, in passing. The table looks terrible from a typesetting point of view, doesn't it? But I have yet to figure out how to typeset LaTeX-style tables in blogger.)
Try generating more such triples, and look for patterns hidden in them. I think there are more patterns there than one may expect.
The same thing is true for NPTs - i.e., triples $(a, b, c)$ of positive integers for which $a^2 + b^2 = c^2 - 1$. Here are a few such triples, generated the same way (with $a \le b \le 30$):
a b c
________
2 2 3
4 8 9
6 18 19
12 12 17
18 30 35
If you experiment on this problem on your own, you will hit upon several oddities which are difficult to explain. For example, there seem to be many more APTs than NPTs. Within the region $a \le b \le 30$, the above two tables list all APTs and all NPTs. Observe that the APTs outnumber the NPTs by a fairly big margin. This discrepancy persists as the upper bound is raised. It is not at all clear why this is so.
We'll continue on this theme in the next post.
We'll continue on this theme in the next post.
The Pythagorean triples can be identified and solved from (p +q)^2 -(p -q)^2 equals 4pq, which has an integer square root if both p and q have integer square roots. comment by Peter L. Griffiths
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