Henry Ernest Dudeney/Puzzles and Curious Problems/168 - Mental Arithmetic/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $168$

Mental Arithmetic
Find two whole numbers (each less than $10$)
such that the sum of their squares, added to their product, will make a square.


Solution

Dudeney gives $3$ and $5$:

$3^2 + 3 \times 5 + 5^2 = 49 = 7^2$

but we also have:

$7^2 + 7 \times 8 + 8^2 = 169 = 13^2$


Proof

Let $a$ and $b$ be the numbers in question.

We have that:

\(\ds a^2 + a b + b^2\) \(=\) \(\ds \paren {a - m b}^2\) for some $m$
\(\ds \) \(=\) \(\ds a^2 - 2 a m b + b^2 m^2\)
\(\ds \leadsto \ \ \) \(\ds b + a\) \(=\) \(\ds -2 a m + b m^2\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds \dfrac {a \paren {2 m + 1} } {m^2 - 1}\)

$m$ can be any integer greater than $1$, and $a$ is chosen to make $b$ an integer.


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The general values are:

\(\ds a\) \(=\) \(\ds m^2 - 1\)
\(\ds b\) \(=\) \(\ds 2 m + 1\)


Setting $m = 2$, we have:

\(\ds a\) \(=\) \(\ds 2^2 - 1\) \(\ds = 3\)
\(\ds b\) \(=\) \(\ds 2 \times 2 + 1\) \(\ds = 5\)


Setting $m = 3$, we have:

\(\ds a\) \(=\) \(\ds 3^2 - 1\) \(\ds = 8\)
\(\ds b\) \(=\) \(\ds 2 \times 3 + 1\) \(\ds = 7\)

and the two solutions are apparent.

$\blacksquare$


Sources

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