638 
SIR FREDERICK POLLOCK ON THE MYSTERIES OE NUMBERS 
For 4w + l+(2w— 1) terms of 2, 4, 6, 8, &c. 
=4m 2 +2w+1 = 
0+i) 2 
which, when expressed in roots, is n, n, n, w + 1. The index of the term is in — 1 
(being the 2wth term of 1, 3, 5, 7, &c.). 
It will be seen that in each case the sum of the roots exceeds the index of the term 
by 2 ; it is therefore obvious that if the sum of the roots could be diminished by 2 
(without altering the sum of the squares), the sum of the roots would be equal to the 
index, and the roots of every term in the diagonal would be obtained, with the sum of 
roots equal to the index of the term ; and as the first number in The Square is in this 
diagonal, the roots of the given number would be obtained, whose sum would equal 1. 
The difference between the sums of the squares of the roots of any two terms in the 
gradation series is the sum of the roots of the larger, minus 1. The difference between 
the sums of their roots is 2 ; therefore, if the sum of the roots of the larger, minus 1, 
can be added to the roots of the smaller, a set of roots will be obtained whose sum will 
be 2 less than that of the larger, but the sum of their squares will be equal to it. (See 
Diagram No. 2.) 
Take the case of 35 as the first term of The Square. In the (w+l)th term of the 
diagonal, 8, 9, 9, 9 will be the roots of the squares which compose that term. The next 
term in the gradation series will have the roots 9, 9, 9, 10, but will be two more 
than the position requires, as will appear from calculation. The difference between 
8 2 + 9 2 + 9 2 + 9 2 and 9 2 + 9 2 +9 2 +10 2 will be 9+9 + 9+10 — 1, that is, 36, but that would 
be 2 too much. If, therefore, the squares of 8, 9, 9, 9 be increased by 34, the number 
will be what that place in The Square requires, and the sum of the roots will be 35, 
which is the index of that term. 
Now the squares of the roots 8, 9, 9, 9 maybe increased by 34 by adding 12 + 184-4, 
which are three terms in the two columns above mentioned ; 12 may be added by 
changing 8, 9 into 6, 11 ; and 18 may be added by changing 9, 9 into 6, 12. The 
roots then are 6, 11, 6, 12, and to these 4 is to be added. This may be done by 
0 4 3 
changing 11, 12 into 10, 13, and the result is 6, 6, 10, 13, whose sum is 35, and is 
therefore equal to the index of the term. 
To reduce roots whose sum is 35 to roots having the same differences but whose sum 
will equal 1, deduct 9 from each root, and 9x4 will be deducted, and the sum will then 
be —1, which by changing all the signs will become +1. From each of the roots 6, 
6, 10, 13 deduct 9, and they will then be —3, —3, 1, 4, and the sum of their squares 
= 35. 
If 1 be deducted from 35, and the remainder be divided by 2, the quotient (17) will 
be composed of not exceeding three trigonal numbers. 
