another zero square Rya,x+by according to No. I, we form a zero magic square. For, we reproduce exactly the N permutations of N things taken in pairs with repetitions. Thus, supposing to be any one element of the latter square, we have in either of the two squares which combines Ryax+oby and Rya+by, the pairs k, n and -n, k; k, and n+1, k; ... k, n and n, k; and after the elements -n, − n + 1, ... n-1, n have all been permuted in the process of forming a zero square from Ryax+oby, all these permutations still exist, as before, only in other situations. - n+1 We cannot of course combine two squares derived from Ryax+oby and Rya,x+oby because N and the coefficients of y have a common measure. Again, by the theorem above it will be seen that if in the zero square formed from Ryax + boy, we substitute Ryv + ke for every element V2 where k is any integer, we still have a zero square. That is, from any zero square for Ryax + boy may be at once derived another for Rxax + σ (by + k), because each of the rows and diagonals still consists of the series-n, - n + 1, n-1, n, in new situations; also by the theorem the zero columns still remain such. But next suppose a+ob to contain a factor p, common to N; then we cannot choose any zero parallelogram of v's at pleasure, but only one so arranged that the period of in the first parallelogram (prolonged N ρ ν ν terms V1227 339 ... by repetition if necessary) shall be a zero period. And further, as the theorem does not apply to such zero periods, we cannot derive the zero square for Ryax+o (by+k) from that for Ryax+boy, as in the former case. The same remarks apply to a ob and the diagonal Ve, 19 Vp-1,27 Examples. (1) R2x+9y, when = 1, 71, gives 420 246 753 113 576 576 420 246 753 113 ν p-2,39 Take any parallelogram of 3 zero columns of 5 terms each formed as above, as 567 234 011 243 567 Then the series of permutations will be 420 5 7 6 1 13 753 24 5 6 7 2 3 4 011 243 567 Hence we derive the zero square for R,,2x + 3 (3y+1) as 15 (2) R1x+by, = 1, n = 1, gives 776 543 210 123 4 5 6 Here the acute diagonal has the period 1 4 6, and therefore in arranging the zero columns, we must provide that this be replaced by a zero period. We take the same arrangement as before, the order of the numbers in the columns being slightly altered, viz. 15 But we cannot derive from this, as in the preceding example, the zero square for R2+3 (2y + k). Thus, if k = 1, we should have in the acute diagonal the non-zero period 4, 2, 5. It is manifest that the method just employed may be applied to the excepted case in No. I., where a+b or a-b or each of them, is commensurable with N; as in R, 2x + 7y; but a numerical example would occupy too much valuable space. 15 For the form Ryoax + by, it is obvious that we have to proceed exactly as for Ryax+oby, only putting rows for columns and columns for rows throughout the process. Suppose next that we have the form Roax +rby, where N=POT, σ and 7 being according to the original assumption prime to each other. The preceding investigation shews that the square resolves itself into po rows recurring 7 times, and that each set of ρστ these rows breaks up in po parallelograms of elements 7 in rank, po in file, thus separating the whole square into por or N parallelograms 7 in rank, po in file, each parallelogram exactly comprising the N elements-n, n+1,...n-1, n. The same investigation shews in like manner that the whole square separates into N parallelograms pr in rank, & in file, each parallelogram comprising the elements-n, -n+1, ... n-1, n. From both these statements together, we infer that the square separates into pN smaller parallelograms, 7 in rank, σ in file; and that p of these parallelograms taken either in row or in column, exactly contain the elements - n, n + 1, ... n − 1, n. We may therefore proceed to the corresponding zero square by forming the series - n, − n + 1, ... n-1, n into either pr zero columns of σ terms each, or po zero rows of terms each. The reader will easily illustrate what has been said upon this case by such an example as R103ax+5by, R103ax+7by, or R15ax+7by. 105 The last two would have partial periods in the grave diagonal. Though the simplest of their kind, they are too long for insertion here, even in the most abridged form. Everton Vicarage, near St. Neots. NOTE ON A SYSTEM OF ALGEBRAICAL EQUATIONS. By Prof. CAYLEY. CONSIDER the system of equations. a+b (y + z)2 + cy*z* = 0, a+b (z+x)2 + cz2x2 = 0, a+b (x + y)2 + cx3y2 = 0, which is a particular case of that belonging to the porism of in-and-circumscribed triangle. We have y and z the roots of a + bx2 + 2u.bx+u2 (b + cx2) = 0. or substituting in the equation between y and z, this becomes (ac+b2) (a + 4bx2 + cx1) = 0, so that if ac + b2 is not = 0, we have so that x=y or else x=z. If x=z, the three equations reduce themselves to the two a+bx2+2y.bx+ y2 (b + cx2) = 0, α + 4bx2 + cx = 0, 3bx + cx* giving y=x, or else y=-; and it hence appears b+cx2 that if from this last equation and a + 4bx2 + cx=0 we eliminate x, the result must be a + 4by+cy=0. For in the same way that the elimination of y, z from the original three equations gives a + 4bx2 + cx = 0, the elimination of x, z from the same three equations will give a + 4by2+cy* = 0, so that in any case y is a root of this equation. |