Mr Penberthy’s Problem

It turns out I was wrong: there is something worse than spurious pseudocontext. It's pseudocontext so creepy it made me throw up a little bit:

Yeah, I know: things were different, 250-odd years ago, but still, ew.

I managed to put my disgust to one side for long enough to solve the thing, though - at least, partly.

We've got:

(1) $x^3 + xy^2 = 4640y$
(2) $x^2y - y^3 = 537.6x$

I'd rewrite them slightly:

(1') $x(x^2 + y^2) = 4640y$
(2') $y(x^2 - y^2) = 537.6x$

Now, what with all of those xs and ys knocking around, it looks sensible to multiply the two together:

(3) $xy(x^2+y^2)(x^2 - y^2) = 464 \times 5376xy$, shuffling a 10 about to make things nicer.

We know that $xy \ne 0$ because the odious Penberthy is old enough to be writing stalkerish poetry, so we can get to $x^4 - y^4 = 464 \times 5376$, using difference of two squares.

Let's factorise the RHS: it's $(2^4\times 29) \times (2^8 \times 3 \times 7)$, which suggests there's a factor of $2^{12}$ to come out of everything. Let $8X = x$ and $8Y = y$, giving $X^4 - Y^4 = 3 \times 7 \times 29$.

The possible factor pairs making that up are:

  • 1 and 609
  • 3 and 203
  • 7 and 87
  • 21 and 29

The last pair are conveniently spaced either side of 25, suggesting $(X^2-Y^2)(X^2+Y^2) = 21 \times 29$, making $X=5$ and $Y=2$. Mr Penberthy is thus 40 and the poor object of his affections 16.

Dude. No. The answer to your question is, mathematically, "Both - you are too old and she is too young." Morally, it's "If you have to ask if it's inappropriate, it's inappropriate."

Another way

The above approach assumed that the answer had to be integers. There is another way.

Let $y = zx$, where $z$ is a constant to be determined.

A bit of work on (1) and (2) leaves us with:

(1'') $x^2(1 + z^2) = 4640 z$ and

(2'') $x^2z(1 - z^2) = 537.6$

Dividing leads to:

$\frac{1+z^2}{z(1-z^2)} = \frac{4640}{537.6}z$

I'll take the opportunity to tidy up that fraction:

$\frac{1+z^2}{z(1-z^2)} = \frac{725}{84}z$

So, $84(1+z^2) = 725z^2(1-z^2)$, which can be rearranged as $725z^4 - 641z^2 + 84 = 0$, and factorised (obviously) as $(25z^2 - 4)(29z^2 - 21)$, giving $z = \pm \frac{2}{5}$ and $z=\pm \sqrt{\frac{21}{29}}$.

Now, $z$ must be positive, from the context. (Ew! I'd almost managed to forget the context), so $z = \frac{2}{5}$ or $z= \sqrt{\frac{21}{29}}$. Let's substitute the first of those into (1''):

$x^2 \times \frac{29}{25} = 4640 \times \frac{2}{5}$, so $x^2 = \frac{4640 \times 2 \times 25}{5 \times 29} = 1,600$, so $x=40$, as before. Ew.

The second one, though, gives:

$x^2 \times \frac{50}{29} = 4640 \times \sqrt{\frac{21}{29}}$, so $x^2 = 4640\times \frac{29}{50} \times \frac{21}{29} \approx 2290.111$, so $x \approx 47.855$ and $y \approx 40.723$. That - while an unlikely answer - would at least be a much healthier age gap.

* Thanks to @aap03102 for the puzzle and discussion.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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