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

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

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.

### One comment on “Mr Penberthy’s Problem”

Awesome post Colin. Thanks for sharing.

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