# James Gregory's manuscripts on algebraic solutions of equations

The following is a version of Herbert Turnbull's description of his discovery of James Gregory's attempts at finding an algebraic method to solve polynomial equations. Gregory's work gave new methods for solving cubic and quartic equations [often called biquadratic equations] which he hoped would generalise to allow quintic equations to be solved.

A perusal of the letters which passed between Gregory and Collins will show that one of the final subjects to be discussed was the theory of equations. Ever since the famous discoveries of solutions for cubic and biquadratic equations by Scipio Ferro, Tartaglia and Cardan, of the Italian school, mathematicians of all countries had attempted a generalization, and particularly addressed themselves to find an algebraic solution of the quintic equation. A noteworthy attempt on the quintic and the sextic was made, for example, by Tschirnhausen (1651-1708), who even solved the problem in the case when the second and third highest terms were absent, publishing his results in the 'Acta Eruditorum' of 1683. This led him to hope for equal success in the general case of the quintic and sextic, but Leibniz warned him against setting his hopes too high. It was not, indeed, until the beginning of the nineteenth century that the matter was settled, when Abel demonstrated the impossibility of such a solution, in general, for the quintic and higher equations.

It now appears, from a study of three unpublished manuscripts of James Gregory, which have found their way respectively to the libraries of the universities of Edinburgh and St Andrews and that of the Royal Society at Burlington House, that the subject attracted the serious attention of Gregory, especially during the last six months of his life. These three manuscripts give a new way of solving the cubic and biquadratic equations, together with a remarkable attempt upon the quintic. A comparison between these notes and the allusions contained in the letters of Gregory and Collins, from May to October, 1675, make it evident that Gregory believed he had hit upon a general algebraic solution for all orders. "I have now abundantly satisfied myself in these things I was searching after in the analytics, which are all about reduction and solution of equations. It is possible that I flatter myself too much, when I think them of some value, and therefore am sufficiently inclined to know others' thoughts," as he writes in May, 1675. A week or two later Tschirnhausen, then a young man of twenty-four years, arrived in London, to spend the summer in a search for the solution of all equations up to the eighth dimension. Collins, who gave a picturesque description of their meeting and conversation, heard of the universal method of Tschirnhausen but only received, apparently, a few trivial examples which he passed on to Gregory, who, in turn, replied in his letter to Collins, dated August 20, in generous terms: "This gentleman's universal method, when it comes to be public, may be more compendious than mine."

Gregory solved the cubic by reducing it rationally to a quadratic, and solved the biquadratic by a corresponding transformation to a cubic. These transformations he called canons: they differ for each dimension, and the labour of carrying them out "increases at a strange rate as the dimensions augment." He went so far as to propose to Collins that the canons of the first ten dimensions be calculated; and, by October, Collins had agreed to persuade Dary to do this, if the necessary rules and demonstrations were communicated to them by Gregory. But the sudden death of Gregory left this project unfulfilled: nor did Collins ever receive particulars of the method. The papers later fell into the hands of David Gregory, the nephew of James, who made some comments on them but never published them.

Evidently James Gregory believed that he could reduce a quintic to a quartic, and, more generally, an $n$-ic to a lower equation. His method depended on a series of eliminations, which he carried out fully for the cubic and biquadratic, but only partially for the quintic. A complete elimination would inevitably have disclosed the need for solving a sextic before the end was achieved, and would have convinced Gregory that his general method failed for $n > 4$. Had Dary, an energetic and skilled computer, embarked upon this enterprise, a highly interesting development in the theory of equations would have followed.

Actually Gregory did not set too great store by this general method: for, as he wrote in his very last letter to Collins, "I have a method of series going very far beyond it, and, to be ingenuous, I think one of its greatest uses is by drawing learned men from this contemplation which busied so many fine wits."

Gregory writes his mathematical argument in Latin. We give a version close to Turnbull's translation.

**Cubic equations:**

Given $x^{3} + q^{2}x + r^{3} = 0$ put $x = z + v$. From this is made the following equation

$v^{3} + 3zv^{2} + (q^{2} + 3z^{2})v + (z^{3} + q^{2}z + r^{3}) = 0.$

Multiply this by the arbitrary cubic $v^{3} + av^{2} + b^{2}v + c^{3}$ to obtain
$v^{6} + (3z + a)v^{5} + (q^{2} + 3z^{2} + 3za + b^{2})v^{4} + (z^{3}+ q^{2}z + r^{3} + q^{2}a + 3zb^{2} + c^{3} + 3az^{2})v^{3} + (z^{3}a + q^{2}za + r^{3}a + b^{2}q^{2} + 3z^{2}b^{2} + 3c^{3}z)v^{2} + (b^{2}z^{3} + b^{2}q^{2}z + b^{2}r^{3} + c^{3}q^{2} +3c^{3}z^{2})v + (c^{3}z^{3} + c^{3}q^{2}z + c^{3}r^{3}) = 0.$

Equate the coefficients of $v^{5}, v^{4}, v^{2}, v$ to 0 to obtain a quadratic in $v^{3}$
(1) $3z + a = 0$,

(2) $q^{2} + 3z^{2} + 3za + b^{2} = 0$,

(3) $z^{3}a + q^{2}za + r^{3}a + b^{2}q^{2} + 3z^{2}b^{2} + 3c^{3}z = 0$,

(4) $b^{2}z^{3} + b^{2}q^{2}z + b^{2}r^{3} + c^{3}q^{2} + 3c^{3}z^{2} = 0$.

If through the first equation $a$ is removed both from the second and the third, the second equation becomes $-q^{2} + 6z^{2} = b^{2}$, and the third
(2) $q^{2} + 3z^{2} + 3za + b^{2} = 0$,

(3) $z^{3}a + q^{2}za + r^{3}a + b^{2}q^{2} + 3z^{2}b^{2} + 3c^{3}z = 0$,

(4) $b^{2}z^{3} + b^{2}q^{2}z + b^{2}r^{3} + c^{3}q^{2} + 3c^{3}z^{2} = 0$.

$-3z^{4} - 3q^{2}z^{2} - 3r^{3}z + b^{2}q^{2} + 3b^{2}z^{2} + 3c^{3}z = 0$.

From these two equations and the fourth are found three values
$b^{2} = 6z^{2} - q^{2},$

$b^{2} = (3z^{4} + 3q^{2}z^{2} + 3r^{3z} - 3c^{3}z)/(q^{2} + 3z^{2}),$

$[-]b^{2} = (q^{2}c^{3} + 3c^{3}z^{2})/(z^{3} + q^{2} + r^{3})$.

From these emerge the following two equations without $b$,
$b^{2} = (3z^{4} + 3q^{2}z^{2} + 3r^{3z} - 3c^{3}z)/(q^{2} + 3z^{2}),$

$[-]b^{2} = (q^{2}c^{3} + 3c^{3}z^{2})/(z^{3} + q^{2} + r^{3})$.

$15z^{4} - q^{4}- 3r^{3}z + 3c^{3}z = 0,$

$6z^{5} + 5q^{2}z^{3} + 6r^{3}z^{2} - q^{4}z - q^{2}r^{3} + q^{2}c^{3} + 3z^{2}c^{3} = 0.$

Hence emerge two values of $c^{3}$,
$6z^{5} + 5q^{2}z^{3} + 6r^{3}z^{2} - q^{4}z - q^{2}r^{3} + q^{2}c^{3} + 3z^{2}c^{3} = 0.$

$-c^{3} = (15z^{4} - q^{4} - 3r^{3}z)/3z,$

$-c^{3} = (6z^{5} + 5q^{2}z^{3}+ 6r^{3}z^{2} - q^{4}z -q^{2}r^{3})/(q^{2} + 3z^{2})$

and hence there results the following equation having no unknown quantity except z,
$-c^{3} = (6z^{5} + 5q^{2}z^{3}+ 6r^{3}z^{2} - q^{4}z -q^{2}r^{3})/(q^{2} + 3z^{2})$

$27z^{6} - 27r^{3}z^{3} - q^{6} = 0$.

**Quartic equations:**

Given $x^{4} + q^{2}x^{2} + r^{3}x + s^{4} = 0$. Let $x = z + v$; hence the following equation

$v^{4} + 4zv^{3} + (q^{2} + 6z^{2})v^{2} + (4z^{3} + 2q^{2}z + r^{3})v +(z^{4} + q^{2}z^{2} + r^{3}z + s^{4}) = 0$.

Multiply these terms by $v^{2} + av + b^{2}$, so that
$v^{6} + (4z + a)v^{5} + (q^{2} + 6z^{2} + 4az + b^{2})v^{4} + (4z^{3} + 2q^{2}z + r^{3} + aq^{2} + 6az^{2} + 4b^{2}z)v^{3} + (z^{4} + q^{2}z^{2} + r^{3}z + s^{4} + 4az^{3} + 2aq^{2}z + ar^{3} + b^{2}q^{2} + 6b^{2}z^{2})v^{2} + (az^{4} + aq^{2}z^{2} + ar^{3}z + as^{4} + 4b^{2}z^{3}+ 2b^{2}q^{2}z + b^{2}r^{3})v + (b^{2}z^{4} + b^{2}q^{2}z^{2} + b^{2}r^{3}z + b^{2}s^{4}) = 0.$

Equate the coefficients of $v^{5}, v^{3}, v$ to 0 to obtain a cubic in $v^{2}$
(1) $a + 4z = 0$,

(2) $4z^{3} + 2q^{2}z + r^{3} + aq^{2} + 6az^{2} + 4b^{2}z = 0$,

(3) $az^{4} + aq^{2}z^{2} + ar^{3}z + as^{4} + 4b^{2}z^{3} + 2b^{2}q^{2}z + b^{2}r^{3} = 0$.

If through the first $a$ is removed from the rest, the following are formed
(2) $4z^{3} + 2q^{2}z + r^{3} + aq^{2} + 6az^{2} + 4b^{2}z = 0$,

(3) $az^{4} + aq^{2}z^{2} + ar^{3}z + as^{4} + 4b^{2}z^{3} + 2b^{2}q^{2}z + b^{2}r^{3} = 0$.

$20z^{3} + 2q^{2}z - r^{3} - 4b^{2}z = 0,$

$4z^{5} + 4q^{2}z^{3} + 4r^{3}z^{2} + 4s^{4}z - 4b^{2}z^{3} - 2b^{2}q^{2}z - b^{2}r^{3} = 0,$

and therefore
$4z^{5} + 4q^{2}z^{3} + 4r^{3}z^{2} + 4s^{4}z - 4b^{2}z^{3} - 2b^{2}q^{2}z - b^{2}r^{3} = 0,$

$b^{2} = (20z^{3} + 2q^{2}z - r^{3})/4z,$

$b^{2} = (4z^{5} + 4q^{2}z^{3} + 4r^{3}z^{2} + 4s^{4}z)/(4z^{3} + 2q^{2}z + r^{3})$

and hence
$b^{2} = (4z^{5} + 4q^{2}z^{3} + 4r^{3}z^{2} + 4s^{4}z)/(4z^{3} + 2q^{2}z + r^{3})$

$64z^{6} + 32q^{2}z^{4} + (4q^{4} - 16s^{4})z^{2} - r^{6} = 0$.

**Quintic equations:**

Given $x^{5} + q^{2}x^{3} + r^{3}x^{2} + s^{4}x + t^{5} = 0$ put $x = z + v, x^{2} = z^{2} + 2zv + v^{2}, x^{3} = z^{3} + 3z^{2}v + 3zv^{2} +v^{3}, x^{5} = z^{5} + 5z^{4}v + 10z^{3}v^{2} + 10z^{2}v^{3} + 5 z v^{4}+ v^{5}$. From these is made the following equation

$v^{5} + 5zv4 + (10z^{2} + q^{2})v^{3} + (10z^{3} + 3q^{2}z + r^{3})v^{2} + (5z^{4} + 3q^{2}z^{2} + 2r^{3}z + s^{4})v + (z^{5} + q^{2}z^{3} + r^{3}z^{2} + s^{4}z + t^{5}) = 0$.

Let this equation be multiplied by the following
$v^{15} + av^{14} + b^{2}v^{13} + c^{3}v^{12} + d^{4}v^{11} + e^{5}v^{10} + f ^{6}v^{9} + g^{7}v^{8} + h^{8}v^{7} + k^{9}v^{6} + l^{10}v^{5} + m^{11}v^{4} + n^{12}v^{3} + p^{13}v^{2} + D^{14}v + q^{15} = 0$.

The resulting equation has 20 dimensions. If all its terms are separately equated to zero, except those of dimension 20, 15, 10, 5, 0, equations are formed, 16 in number. But there are 17 unknown quantities (namely $v, z, a, b,$ etc.); and since the resulting 16 equations and the one given make 17, it is clear that with the help of these operations all the unknown quantities can be removed except any one namely $v$. If this is done, a value of $v$ will be given by a biquadratic equation having a quintic root: and so the value of $v$ will be given with the help of a biquadratic equation, all whose terms are explicit, and a pure quintic.
Also the value of $z$ will be given in rejecting the unknown quantities; namely when $a$ is rejected a duplicate value $5z$ is found for the same, which can more easily be picked out. Therefore $x = z - v$ is given since both $z$ and $v$ are made known: and hence it is clear in what manner the quintic equation can be reduced to a biquadratic, if only someone may be found who would not recoil from such labour.

$v^{20}+ (5z + a)v^{19}+ (10z^{2}+ q^{2}+ 5az + b^{2})v^{18}+ (10z^{3}+ 3q^{2}z + r^{3}+ 10az^{2} + aq^{2} + 5b^{2}z + c^{3})v^{17} + ... = 0$

[Gregory actually writes out this equation in full. We chose to give up here!]
Equate the coefficients of $v^{19}, v^{18}, v^{17}, v^{16}, v^{14}, v^{13}, ...$ to 0 to obtain

$[-]a = 5z$

$b^{2} = 15z^{2} - q^{2}$

$[-]c^{3} = 35z^{3} - 7q^{2}z + r^{3}$

$d^{4} = 70z^{4} - 163q^{2}z^{2} + 8r^{3}z - s^{4}+10q^{4}$

$f ^{6} = -420z^{6} + 1560q^{2}z^{4} - 60v^{3}z^{3} + 45q^{4}z^{2} - 10q^{2}r^{3}z + 5t^{5}z + 2q^{2}s^{4} + r^{6} - 10q^{6} - 5ze^{5}$

$g^{7} = -1580z^{7} +$ etc. [Gregory gives 14 terms]

$h^{8} =$ [Gregory gives the whole expression]

$b^{2} = 15z^{2} - q^{2}$

$[-]c^{3} = 35z^{3} - 7q^{2}z + r^{3}$

$d^{4} = 70z^{4} - 163q^{2}z^{2} + 8r^{3}z - s^{4}+10q^{4}$

$f ^{6} = -420z^{6} + 1560q^{2}z^{4} - 60v^{3}z^{3} + 45q^{4}z^{2} - 10q^{2}r^{3}z + 5t^{5}z + 2q^{2}s^{4} + r^{6} - 10q^{6} - 5ze^{5}$

$g^{7} = -1580z^{7} +$ etc. [Gregory gives 14 terms]

$h^{8} =$ [Gregory gives the whole expression]

**Explanation.**

Gregory takes a cubic $f (x) = x^{3} + q^{2}x + r^{3}$, from which the second term is absent, and later a similar biquadratic and quintic. The exponents of the coefficients accord with his usual idiom, rendering the results homogeneous and virtually acting as the weight-suffixes of Cayley and Sylvester. Next Gregory puts $x = z + v$ and multiplies the resulting cubic in $v$ by an arbitrary cubic $y(v) = v^{3} + av^{2} + b^{2}v + c^{3}$. (For the biquadratic he takes $y(v)$ to be an arbitrary quadratic, and, for the quintic, a polynomial of degree 15.) The resulting sextic in $v$ he majestically arranges in full, and then causes the coefficients of $v^{5}, v^{4}, v^{2}$ and $v$ to vanish, so that the sextic becomes a quadratic in $v^{3}$. This gives him four equations for $z, a, b, c$ in terms of the known $q$ and $r$, from which he eliminates $a, b, c$ in succession, and obtains the relation

$27z^{6} - 27r^{3}z^{3} - q^{6} = 0$.

In the St Andrews manuscript there is an error in sign in the last equation for $b^{2}$ and in the next for $c^{3}$, but the final result is correct. In the Gregory volume at Edinburgh, which contains a duplicate note in his handwriting, the signs have been corrected, and an additional remark is added, that
$a = -3z, b^{2} = 6z^{2} - q^{2}, c^{3} = r^{3} - 5z^{3} + q^{4}/3z$.

It is easy to verify that the assumption $x = q^{2}/3z - z$ leads to this quadratic for $z^{3}$ in place of the original cubic in $x$.
For the biquadratic Gregory builds up a sextic in $v$, involving three unknowns, $z, a, b$ where $x = z + v, y(v) = v^{2} + av + b^{2}$. By equating coefficients of $v^{5}, v^{3}, v$ to zero he obtains three relations, whence $a$ and $b$ are eliminated and a cubic in $z^{2}$ remains:

$64z^{6} + 32q^{2}z^{4} + (4q^{4} - 16s^{4})z^{2} - r^{6} = 0.$

There follow a few indistinct words in the St Andrews manuscript suggesting that, from these values of $z^{2}$, values of $v^{2}$ and of $x$ can be found.
These notes on the cubic and biquadratic occur practically in duplicate upon two sheets of foolscap paper, one in the St Andrews manuscript and one in the Edinburgh manuscript, the former occupying the back of a letter from Collins. On the back of the latter is a further note .... In the London manuscript, alone, the attempt on the quintic appears, elaborately written out across the double page. Some comments by David Gregory, the nephew of James Gregory, occur on the back of this sheet.

The method of dealing with the quintic is clear from Gregory's own statement. He multiplies $F(v)$, his quintic in $v$, by an arbitrary quantic $y(v)$ of order 15 so as to get the order 20, which is the L.C.M. of 5 and 4, and thereby, after specializing the coefficients, to produce a biquadratic in $v^{5}$. Unfortunately the elimination processes would have yielded at least one irreducible equation of degree 6, had the attempted work been completed: and thus the hoped-for solution - by lowering the degree from 5 to 4 - would have had to be abandoned. It may also be remarked that, to complete the theory of the cubic and biquadratic, it would be necessary to show that the values of $z$ and $v$, so found, would yield roots of the original equation, and not merely those of the arbitrary factor $y(v)$. Doubtless Gregory would have done this, had he lived to publish the work. The whole matter reveals the workings of a mind possessing deep insight into polynomial structure. One cannot but be impressed with the breadth and vigour of the conception, a veritable unfinished symphony, and a true forerunner of the Cayley, Salmon, Sylvester developments in the nineteenth century.

In a Latin note David Gregory comments on this proposed general method of his uncle. He asks two questions:

(1) How is the degree of the arbitrary multiplying polynomial determined?

(2) Why is the degree in $z$ of the final eliminant lower than that of $x$ in the original equation?

No doubt a satisfactory reply to (1) could be made, but, as for (2), it is insuperable. The degree for the quintic and further cases would be higher, and not lower.
(2) Why is the degree in $z$ of the final eliminant lower than that of $x$ in the original equation?

On the back of the Edinburgh sheet are a few further notes by James Gregory on the connexion between solving quadratics and ruler and compass constructions.

Presumably Gregory was discussing the problem of drawing a straight line through a vertex of a given square so as to intercept a given length $b$ on the line between the two sides remote from the vertex. At any rate he readily gets the equation

$z^{4} - 2ayz^{2} - b^{2}y^{2} = 0$,

which he factorizes as $(z^{2} - ty)(z^{2}+ vy) = 0$.
This gives $v = c - a, t = c + a$ where $c = √(a^{2} + b^{2})$.

But $z^{2} = a^{2} + y^{2}$: hence

$y^{2} ± (c - a)y + a^{2} = 0$.

Thus the problem which apparently led to a biquadratic equation is, in fact, soluble in quadratics. As to this Gregory added a significant remark, which is here rendered from the Latin into English:
In this way we see that whenever a biquadratic equation lacks its second and fourth term, but contains one unknown quantity to one dimension [y] in the third and to two [$y^{2}$] in the last term, it is possible to eliminate the unknown which occurs to four dimensions [$z^{4}$], and thereby to resolve the said biquadratic into two quadratics, where the unknown quantity is confined to two dimensions. All problems therefore which can be reduced to this are soluble by plane geometry: yet a sure method is wanting whereby in every possible case problems may be reduced to this.

Last Updated July 2012