Consecutive Pythagorean triangle sides

In this post we find all Pythagorean triples that contain consecutive numbers, all Pythagorean triples (a, b, c) such that a + 1 = b or b + 1 = c.

a + 1 = b

George Osborne wrote a paper [1] addressing the question of when the squares of two consecutive numbers is also a square. Geometrically this is asking for primitive Pythagorean triples for which the legs are consecutive integers.

He proved that the sequence shorter legs satisfies the recurrence relation

$u_{n+2} = 6 u_{n+1} - u_{n+1} + 2$
with initial conditions u₀ = 0 and u₁ = 1. This is OEIS sequence A001652.

The method for solving recurrences like the one above is analogous to the method for solving linear differential equations. See a solution here. This gives us the following formula for the terms:

$u_n = \dfrac{1 + \sqrt{2}}{4} \left(3 + 2\sqrt{2}\right)^n + \dfrac{1 - \sqrt{2}}{4} \left(3 - 2\sqrt{2}\right)^n - \dfrac{1}{2}$

b + 1 = c

It’s also possible for the longer side and hypotenuse of a Pythagorean triangle to be consecutive numbers, as in the (5, 12, 13) triangle.

All primitive Pythagorean triples are given by Euclid’s formula

$\begin{align*} a &= m^2 - n^2 \\ b &= 2mn \\ c &= m^2 + n^2 \end{align*}$

with integers m > n > 0. If b and c are consecutive numbers, then

and so m = n + 1. Therefore all possible values of b are given by 2n(n + 1) for n > 1.

[1] Geo. A. Osborne. A Problem in Number Theory. The American Mathematical Monthly, Vol. 21, No. 5 (May, 1914), pp. 148-150

a + 1 = b

b + 1 = c

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