Consider the curve

6nx*y2+12rz2xy + (4gx + 4iy+cz) = 0,

where x=0 is the tangent at a cusp; y=0 the tangent at a cusp; and z = 0 the line joining the two cusps. For the special discriminant we have

3nxy2+3ryz2+gz3 = 0,

3nx3y + 3rxz2 + iz3 = 0,

z {6rxy + (3gx + 3iy + 4cz) z} = 0;

the last of which may be replaced by the equation of the curve. Assume x=Xz, y = uz, the first two equations give

whence also

3 (nλμ + r) μ + g=0,

3 (nλμ + r) λ + i=0,

6nλ3μ3 +6rλμ+gλ+iμ=0,

and the equation of the curve gives

6ηλ με + 12λμ + 4gλ + 4iu + c=0,

whence eliminating gλ + iμ we find

18ηλ με + 12 λμ - c = 0.

Moreover the first two equations give

or substituting in 18n0+ 12r0-c=0, this is

8n (3ig-2cr)+8r (3ig - 2cr) (2r2 + cn) - c (2r2 + cn) = 0. Hence developing, the special discriminant is

0=-1 c3n2

+12 c2nr

-72 cginr

-36 cr

+72 g'i'n

+48 gir3,

which is as it should be of the degree 5,3.3-11.2.

ON THE ALGEBRA OF MAGIC SQUARES. No. III. By JOSEPH HORNER, M.A., Clare College.

HAVING already discussed the various classes of odd magic squares connected with the form Ryax+by, we now proceed to even squares. In doing this, we shall represent a series of unities by their signs only, and in this sense speak of multiplying by those signs.

Take in any order the zero arithmetical series of 2n elements, 2n-1, -2n+1, ... 2n-1, 2n+1, and let λ and -λ represent any conjugate pair of these elements. If λ and -λ be each multiplied by one and the same series of 2n signs, and a like operation extended to all the values which can take, and the results be arranged in columns

according to the assumed order of the original series, we manifestly form a square, every row of which is a zero series.

-

th

We have next to consider what must be the nature of these series, in order that the diagonals and columns of the square may be zero series. Here simplicity will be promoted by taking axes through the centre of the square parallel to its sides, and numbering the distances of the rows and columns from the centre positively and negatively on these coordinates as in the Cartesian system. Then (x, y) will denote a point at the intersecton of the xth row and y' column. Suppose then that the conjugate columns headed with λ and are the 7th and sth. The grave diagonal will meet these at the points and s respectively, reckoned from the axis of x. If, therefore, maintaining the principle already laid down for the zero rows, we multiply λ and -λ by the same sign (which we may express by the optional sign a), at the points and s of both columns, λ and X will both appear in the grave diagonal; and when a similar operation has been extended to all the values that can take, the grave diagonal will be a zero series. Again, the acute diagonal meets the same columns at -r and -s respectively; and if at these points we multiply by the optional sign b, and apply the like process to all the values of X, the acute diagonal will be a zero series. We may now so choose the signs a and b, and those for the remaining 2 (n-2) vacant places of each pair of conjugate columns, that n of them shall be positive and n negative. Thus the conjugate columns of X's are made zero series; and when this has been done for every value of λ, the zero square is completed. If, however, for any particular value of X, the elements are equidistant from the extremities of the series in its assumed order, i.e. if r+s=0, then - s=r, and −r=s; so that the columns are symmetrically situated and the two sets of signs a, b, coalesce into one set. In this case we can choose the sign a and those in the 2 (n-1) remaining places in the same manner as before stated. It will be convenient to keep the sign-columns separate from the terms of the arithmetical series, and to speak of this system as the sign-square. If n = 2 or 3, it is manifest that a and b must represent opposite signs. Also, if n be even, the 2(n-2) or 2 (n - 1) vacant places can always be filled up, if desirable, with pairs of like signs; but if n be odd there must be at least one pair of unlike signs in every column. In the following examples each letter represents an optional sign. The dots stand in

vacant places which are to be afterwards filled up with

suitable signs:

It will be seen that, while we keep the conjugate columns of the sign-square unaltered, the numbers, 1, 3, 5, ... may be interchanged in any manner.

Also, reversing the order of all the columns would cause no change in any of them as to signs.

We now go on to enquire how and under what conditions two such zero squares can be combined by juxta-position into a magic square.

Suppose A andλ to stand at the head of two columns in the first square, and μ and -μ of two in the second. Turn the second square so that its rows become columns; in particular, so that the upper row becomes the left-hand column. When the squares now have their respective elements in juxtaposition, suppose that the four permutations in which A precedes, standing thus:

17

where 1, λ; M1, M, are numerically λ and μ, but with signs at present unsettled. First imagine A, A to result from multiplying by unlike signs, and μ,,, from multiplying μ by unlike signs. Whence λ=-λ, and μ➖➖μ,; so that the four pairs become A, A,,,,,, which are not all different, whatever the signs belonging to λ and μ,. Next conceive A, A, to result from λ and μ,,, from μ by multiplying in each instance by like signs. Then_λ = X1 and ; and we have the four pairs A, A, A, A‚μ‚; which are all different. Thirdly, if either pair, as A, result from multiplying λ by like, and the other μ from multiply μ by unlike signs, we have λ=λ, and μ-μ1 and accordingly, λμ1, λμ1, λμ all different. Upon the whole, therefore, we shall always obtain the four permutations in which λ stands first, provided λ and μ are not both multiplied by unlike signs in the two places where they stand in their respective columns. And this holds good when λ and are equal.

When λ and are unequal, the juxta-posed squares will contain four other permutations of X and having μ first, to which the same law is applicable; and the eight make up all the permutations of the elements λ, - λ, μ, -μ, which are required to be found in the magic square. When λ and u are equal, they can only have the four permutations above mentioned.