FACTORS OF ANY GIVEN NUMBER. 
15 
It is evident that x must be either equal to, or greater 
than \/ a. If « is a complete square then ofi = a and 
= 0. One pair of factors being in this case and Vff ; 
but if « is not a complete square we take the next whole 
number above a/ a and call it b. 
Square b and deduct a, that is, deduct a from the next 
square above the given number. 
Then if 6^ — a is a complete square we have /\/b'^ — a = y 
and b z= X. 
If b'^ — a is not a complete square we try {b + 1)^ — a, 
{b + 2Y — a, and so on, until we find a remainder, that is, 
a square. 
Let b"^ — a = c. 
Now if we were obliged to square successively b, b -\- 1, 
5 + 2, &c., and deduct a from each, the process would be 
long and troublesome. 
Fortunately there is no need for this, because (5 + 1)^ is 
greater than b"^ by 26 -f- 1, therefore {b -j- 1)^— a = c (26-t- 1) ; 
and (6 + 2)2 is greater than (6 + 1)^ by 2 (6 + 1) +1 =26 + 3, 
therefore (6 + 2)2~a = <? + (26 + 1) + (26 + 3) and so on ; so 
that we get the series of differences we require by adding 
26 + 1, 26 + 3, 26 + 5 and so on to the first difference. Each 
being added to the sum of all that precede it. 
The next operation is to test these differences to see if 
any are perfect squares. 
In doing this we are first helped by the fact that all 
squares are of the form 5 a; or 5 + 1, so that we may 
reject all numbers ending with the figures 2, 3, 7, 8, and to 
these may be added 0, unless there are two, as all square 
ending in 0 must be multiples of 100. 
Again all squares are of the form 4 w or 4 n + 1. In 
applying this test, we need only look at the last two figures. 
If the last figure but one be even, then, for the number to be 
a multiple of four, the last figure must be 0, 4, or 8 ; therefore 
