British Museum, printed on narrow slips of paper, and evidently intended to be pasted pro bono publico in conspicuous situations. We have little doubt that the celebrated problem, generally known as Colonel Titus's problem, was originally proposed in this manner. We have already intimated that this problem is attributed to the wrong person^[1], and we have since discovered a note in MS. Birch, 4411, which expressly states that it was "put by Colonel Titus, who had received it from Dr. Pell." The problem in the most general form is as follows:

${\displaystyle a^{2}+bc=\alpha }$	${\displaystyle (1)}$	${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \ \end{matrix}}\right\}\,}}$
${\displaystyle b^{2}+ac=\beta }$	${\displaystyle (2)}$		to find ${\displaystyle a}$, ${\displaystyle b}$, and ${\displaystyle c}$.
${\displaystyle c^{2}+ab=\gamma }$	${\displaystyle (3)}$

Collins has given a solution which occupies fourteen closely written folio pages, and the more modern solutions have not been comprised in a much shorter compass. Wallis's solution is in the same manuscript. Pell, however, criticises Collins's solution very severely, and ridicules him for not observing that the roots will admit both of positive and negative values.

The problem is generally given with numerical values for ${\displaystyle \alpha }$, ${\displaystyle \beta }$, and ${\displaystyle \gamma }$, and the only possible chance of a short solution is a case in which these numbers bear some definite relation to each other, so as to obtain an equation independent of the given quantities. For instance, Pell gives one wherein ${\displaystyle \alpha =15}$, ${\displaystyle \beta =16}$,