Henry Ernest Dudeney/Puzzles and Curious Problems/203 - A Triangle Puzzle/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $203$

A Triangle Puzzle
In the solution to our puzzle No. $162$, we said that:
"there is an infinite number of rational triangles composed of three consecutive numbers like $3$, $4$, $5$, and $13$, $14$, $15$."
We here show these two triangles.
Dudeney-Puzzles-and-Curious-Problems-203.png
In the first case the area ($6$) is half of $3 \times 4$,
and in the second case, the height being $12$, the area ($84$) is a half of $12 \times 14$.
It will be found interesting to discover such a triangle with the smallest possible three consecutive numbers for its sides,
that has an area that may be exactly divided by $20$ without remainder.


Solution

The triangle whose sides are $2701$, $2702$ and $2703$ has an area of $3 \, 161 \, 340$.


Proof

Extend the following table as you like:

$\begin{array} {rrrrr}

n & p_n & q_n & \text{Height} & \text{Area} \\ \hline 1 & 2 & 4 & 3 & 6 \\ 2 & 8 & 14 & 12 & 84 \\ 3 & 30 & 52 & 45 & 1170 \\ 4 & 112 & 194 & 168 & 16 \, 296 \\ 5 & 418 & 724 & 627 & 226 \, 974 \\ 6 & 1560 & 2702 & 23490 & 3 \, 161 \, 340 \\ \end {array}$

This table is governed by the recurrence relation:

$p_n = \begin {cases} 0 & : n = 0 \\ 2 & : n = 1 \\ 4 p_{n - 1} - p_{n - 2} \end {cases}$

and:

${q_n}^2 = 3 {p_n}^2 + 4$


Each row describes a triangle $T$ whose sides are consecutive integers $q - 1$, $q$, $q + 1$, such that:

the height of $T$ is equal to $\dfrac {3 P} 2$
the area of $T$ is equal to the height multiplied by $\dfrac q 2$.


This is demonstrated in Approximations to Equilateral Triangles by Heronian Triangles, except it's not.


This theorem requires a proof.
In particular: prove, and indeed derive, the above. It's just another Pellian equation, of which I'm terminally bored.
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Sources

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