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 $10$)
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.
This article, or a section of it, needs explaining. In particular: Explain why the general values below are what they are based on the general rule just derived. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $168$. -- Mental Arithmetic
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $233$. Mental Arithmetic