triangles with sides consecutive integers & area integer

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triangles with sides consecutive integers & area integer

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Source: Nordic Mathematical Contest 1995 #4

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#1 h Oct 4, 2017, 11:01 PM • 1 Y

Y by Adventure10

Show that there exist infinitely many mutually non- congruent triangles $T$ , satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer.

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#2 h Oct 5, 2017, 10:16 AM • 1 Y

Y by Adventure10

This is a nice Pell's Equation result.

Let the triangle have side lengths $s-1, s, s+1$ , with $s \geq 3$ . By Heron's formula, the area of the triangle is $\frac{s}{4} \sqrt{3s^2 - 12}$ . In order for this to be an integer, we must have $3s^2 - 12 = k^2$ for some integer $k$ . Using $\pmod{4}$ , we get that both $s$ and $k$ are even (If $s$ is odd, then $3s^2 \equiv 3 \pmod{4}$ , which cannot result in a quadratic residue of $4$ ). Thus, $sk$ is divisible by $4$ , so if $k$ exists, then $\frac{s}{4} \sqrt{3s^2 - 12}$ is an integer.

Further note that $3 \mid k$ since $3 \mid 3s^2 - 12$ . Thus, let $t = \frac{s}{2}$ and $l = \frac{k}{6}$ . Substituting these values in gives us the Pell equation $t^2 - 3l^2 = 1$ . This has fundamental solution $(t_0, l_0) = (2, 1)$ . Thus, using the recursive formula $(t_{n+1}, l_{n+1}) = (t_0 t_n + D l_0 l_n, t_0 l_n + l_0 t_n)$ , we get the following solutions: $(2, 1), (7, 4), (97, 56), \ldots$ . These give the side lengths triplets of $(3, 4, 5), (13, 14, 15), (193, 194, 195), \ldots$ .

Thus, proven.

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