Dudeney's "Puzzles and Curious Problems": 55 : Eliza's Surname

The problem is Problem 55 from

Puzzles and Curious Problems
by
Henry Ernest Dudeney
Thomas Nelson and Sons, Ltd.
London, Edinbourgh, New York, Toronto, and Paris

published 1931, posthumous.

55.— ELIZA’S SURNAME [...] [Mr. ]Smith, [Mr. ]Brown, and [Mr. ]Robinson have provided themselves with a penny pencil each, and took their wives to a stockbroker's office to buy shares. Mary bought 50 more shares than [Mr. ]Brown, and [Mr.]Robinson 120 more than Jane. Each man paid as many shillings per share as he bought shares, and each wife as many pence per share as she bought shares, and every man spent one guinea more than his wife. What was Eliza's surname ?

Annotations
"Smith, Brown, and Robinson have provided themselves with a penny pencil each," means that each of these men bought a pencil for one penny. 12 pennies are 1 shilling, 21 shillings are 1 guinea.

In the same book we also can find the solution:

55.— ELIZA'S SURNAME Obviously we have to take into consideration the three penny pencils, and the puzzle cannot be solved without doing so. Then every man spends (on shares) 251d. more than his wife. Eliza’s surname must have been Robinson, and the transactions as follows:
Brown, 21 ; Mrs. Brown (Mary), 71.
Smith, 49 ; Mrs. Smith (Jane), 169.
Robinson, 289 ; Mrs. Robinson (Eliza), 1,001.

If m is the number shares a man bought and w is the number of shares his wife bought, then the following integer equation must be fulfilled

$$12\cdot m^2=w^2+251 \tag 1$$

To solve this equation one can use https://www.alpertron.com.ar/QUAD.HTM Note that there is also a show steps button. We set $a=12,c=-1,f=-251$ and the remaining parameters to $0$. We get the following output

$12 ⁢x^2 - y^2 - 251 ⁢ = 0$

$x = 5$
$y = 7$

and also:
$x = -5$
$y = -7$

and also:
$x = -5$
$y = 7$

and also:
$x = 5$
$y = -7$

Recursive solutions:

$x_{n+1} = 7 ⁢x_n + 2 ⁢y_n$
$y_{n+1} = 24 ⁢x_n + 7 ⁢y_n$

and also:

$x_{n+1} = 7 ⁢x_n - 2 ⁢y_n$
$y_{n+1} = - 24 ⁢x_n + 7 ⁢y_n$

Written by Dario Alpern. Last updated on 26 November 2021.

\begin{array}{ccrcr} x_{n+1} &=& 7 ⁢x_n &+& 2 ⁢y_n\\ y_{n+1} &=& 24 ⁢x_n &+& 7 ⁢y_n \end{array}

\begin{array}{ccrcr} x_{n+1} &=& ⁢7x_n &-& 2 ⁢y_n\\ y_{n+1} &=&-24 x_n &+& 7 ⁢y_n \end{array}

From this we create the following lists of solutions for $12m^2=f^2+251$

$$\begin{array}{cc}5 & 7\\ 49 & 169\\ 681 & 2359\\ 9485 & 32857\\ 132109 & 457639\\ 1840041 & 6374089\\ 25628465 & 88779607\\ 356958469 & 1236540409\\ 4971790101 & 17222786119\\ 69248102945 & 239882465257\\ \ldots&\ldots \end{array}$$

and

$$\begin{array}{cc}5 & -7\\ 21 & 71\\ 289 & 1001\\ 4025 & 13943\\ 56061 & 194201\\ 780829 & 2704871\\ 10875545 & 37673993\\ 151476801 & 524731031\\ 2109799669 & 7308560441\\ 29385718565 & 101795115143\\ \ldots&\ldots \end{array}$$

The solutions generated form other possible combinations of recurrence relations and start values give only lists of negative numbers and therefore can be ignored.

You can do all these calculations by hand to get these values, I think, and use www.alpertron.com to advice you how to do this, but of course this is very laborious.

Even if we use these algorithm we did not show that these are the only solutions. Maybe there are larger numbers in these tables, where Mary is not Mrs. Brown, where these requirements are fulfilled. I don't think so but I do not want to investigate this further.

So, how could this puzzle be solved at the time when it was published first, without the help of a computer or handheld calculator?

The OP already found the solution $(5,7)$ by trial and error. All the four tuples $(\pm5, \pm7)$ are solutions. One can check the pairs below $100,100$ and find $(21,71)$. Some modular calculations can facilitate this.

Note that

$$502=\det\begin{pmatrix}5 & -7\\ 21 & 71\end{pmatrix}= \det\begin{pmatrix}21 & 71\\ 289 & 1001\end{pmatrix}= \det\begin{pmatrix}289 & 1001\\ 4025 & 13943\end{pmatrix}$$

So given a solution pair we can use the determinant equation to get a linear relation between $m$ and $w$, e.g.

$$\det\begin{pmatrix}4025 & 13943\\ m & w \end{pmatrix}=502$$

that we can substitute in equation $(1)$ and we get a quadratic equation in one variable that we can solve.

Also the following equation holds

$$70=\det\begin{pmatrix}5 & -7\\ 5 & 7\end{pmatrix}= \det\begin{pmatrix}21 & 71\\ 49 & 169\end{pmatrix}= \det\begin{pmatrix}289 & 1001\\ 681 & 2359\end{pmatrix}$$

So in a similar way as before we can get from the value pair in one table the corresponding value pair in the other table.

So from $(\pm5,\pm7)$ and $(21,71)$ we are already able to calculate both tables.

But for this special puzzle we have luck because there is an easier way we can do. When I investigated if Mary has more or less shares than the wife of Mr. Brown I had to solve the equation $(1)$ where I replaced the Mr. Brown's number of shares $m$ by $w-50$, so I get

$$12(w-50)^2=w^2+251$$

This equation has the solutions $$w=\frac{419}{11}\operatorname{,}w=71$$ $w=71$ is an integer solution and the corresponding $m=w-50$ is an integer solution too. So we get $(21,71)$. Also when I investigated if Jane had more or less shares then the wife of Mr. Robinsons I replaced in $(1)$ Mr. Robinsons's number of share $m$ by $w+120$ and solving the resulting equation

$$12(w+120)^2=w^2+251$$

has the solutions

$$w=-\frac{1021}{11}\operatorname{,}w=-169$$

and from this we can also retrieve the integer solution $(-49, -169)$ and therefore $(49, 169)$ form $(1)$. Now we check the possible value for Mr. Robinsons number of shares, $169+120$. Substituting $289$ for $m$ in $(1)$ gives $w=1001$ and we are done. These numbers satisfy the given problem. We did not check if this is the only possible solution. With this numbers Eliza's is the wife of Mr. Robinsons.

All in all I do not know which solution way the author of this puzzle had in his mind, but I think it was the last one, because it is possible to create a similar puzzle without these coincidences.

I have no solution but I start with some observations here as community wiki.

From $$12(\text{shares}(\text{MrRoberts}))^2=(\text{shares}(\text{MrsRoberts})^2+251$$ follows $$12(\text{shares}(\text{Jane})+120)^2=(\text{shares}(\text{MrsRoberts})^2+251$$ and further $$\text{shares}(\text{Jane})<\text{shares}(\text{MrsRoberts})$$ so Jane is not Mrs. Roberts.

If we are able to show that $Mary$ also isn_t Mrs. Robinson then we can conlude that Eliza is Mrs Robinson and the question of this puzzle is solved. But I was not able to show this without calculating first the values of alls shares. This is done in a different answer and I stop.