## Damsel And Suitor

### April 21, 2017

Chris Smith tweets mathematical curiosities at @aap03102. This one caught my eye the other day:

This is from 1779: a time when puzzles were written in poetry, solutions were assumed to be integers and answers could be a bit creepy:

Questions proposed in 1779, and answered in 1780.I. QUESTION 742,

by Mr. John Penberthy.I’m in love with a damsel, the pride of the plain,

Have courted and talk’d in Ovidian strain;

But vain is the rhetoric us’d by my tongue,

She says I’m too old and that she is too young:

From the foll’wing equations, dear ladies, unfold;

If she be too young, or if I be too old.

x^{3}+xy^{2}= 4640y

x^{2}y−y^{3}= 537.6xWhere

xrepresents my age, andythe damsel’s.

Your task is to compute the ages of the damsel and her suitor. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

If you multiply the two left hand equations and the two right hand equations this reduces to: x^4-y^4 = 2494464 or x^4 = y^4 + 2494464

using this we can re can loop through values of y from 1..100 and look for results for which x is an integer…

Solving it 1779 style: the second equation tells us that that x is a multiple of 5, the first, for a given value of x, is just a quadratic in y with the relevant solution being 2320 – sqrt(5382400 – x^4))/x, Mr Napier’s logarithms tell us that the 4th root of 5382400 is 48.1 or thereabouts, so the oldest the man can be is 45. This doesn’t give a correct value for y, so we try the next one down and find that 40 satisfies both equations and gives 16 for the young lady’s age. Smaller values of x rapidly become increasingly unplausible.

Since both persons are most likely between 10 and 100 years of age…. here’s a simple brute-force way

x = 40

y = 16

Using the same logic as James Curtis-Smith, the minimum value of x and y are as below. These values turn out to be the solution.

It is convenient to substitute x=a*y. Excluding the possibility y=0, we can divide through by y. Both equations now only have y^2, which can be eliminated to give the equation

84*a^4-641*a^2+725=0.

This can be solved for a^2 using the quadratic formula. There are two positive solutions for a. Substituting the rational one (a=5/2) into one of the previous equations gives y^2=256 or y=16 and x=40.

The other solution gives y=40*(1.074276)^(1/4), x=y*(1+8/21)^(1/2) or approximately y=40.7 and x=47.8.

The latter ages are not integer, but not creepy.

Several solutions. Mathematica:

Solve[

{

x^3 + x y^2 == 4640 y,

10 x^2 y - 10 y^3 == 5376 x

},

{x, y}

] ~ Cases ~ {

x -> xi_Integer /; xi > 0,

y -> yi_Integer /; yi > 0

}

C11:

#include <iso646.h>

#include <stdbool.h>

#include <stdint.h>

#include <stdio.h>

#include <stdlib.h>

int main(int argc, char **argv) {

const uint32_t maxAge = 200;

puts("Solutions:");

for (uint32_t x = 1; x < maxAge; ++x) {

for (uint32_t y = 1; y < x; ++y) {

if (x*x*x + x*y*y == 4640*y and

10*x*x*y - 10*y*y*y == 5376*x) {

printf(" [ x = %d y = %d ]\n", x, y);

}

}

}

`exit(0);`

}

@Jan: nice solution. Here is the original (from “The Ladies’ Diary”) on Google books:

https://books.google.co.uk/books?id=VPQ3AAAAMAAJ&pg=PA45

The second solution, from Mr. Wm. Reynolds, is like yours. The solution from Mr. Tho. Truswell I don’t entirely follow.

I just brute forced it. In C++.

Which gave:

x = 40 & y = 16