88 On the Analysis of Square Numbers. 
Remarks. 1, When neither a and 6, nor ¢ and d, are equal to 
each other, it is evident that the product A’B contains the sum of 
two squares in two different ways. 
II. When a=8, both powers are reduced to one; viz. 
- A‘B=(ae+ad)?+(ad—ac)?. 
‘III. When b=0, both forms are reduced toone ; viz. 
A:'B=(ac)?+-(ad)’. 
IV. When a=c, we have A°‘B=(a?+6d)? i ab)?. 
and A-B=(a? —bd)? +(ad+ab)? ; 
expressions which still afford two different sets of squares. 
‘V. If a=c, and b=d, the expression [A] becomes (a?-+467)?= 
(a? +b?)2-+-(ad—ad)? =(a?+6")?+(O)?; while [B] assumes 
the form, given by Souri, Bonnycastle, and others, for finding two 
square numbers, whose sum shall be a square: viz. (a7+b?)?= 
(a? —b?)?+(2ab)?. 
VI. If a=b, and e=d, both expressions assume an identical form, 
(2ac)* =(2ac)*, from which nothing can be determined. 
VII... If a=1, and b=1, both expressions [A] and [B] take this 
form, 2(c? +-d?)=(ce+d)*+(c—d)?. . Whence it appears that if a 
given number contain the sum of two squares, the double of that 
number will also contain the sum of two squares, but generally only 
in one wa 
Vill» Finally, if a*+b*=c?-+-d?, while neither the values of a 
and 6, nor of ¢ and d, are alike, we have the expressions, 
A-B=A? =(a? +52)? =(ac+bd)?+(ad—bc)?, and 
(a *-+6?)?=(ac—bd)? +(ad-+be)? 5 that is, a 
square number that contains the sum of two squares in two different 
ways. 
Note.—The condition a?-+b?=c?-+-d?, is satisfied at once by 
substituting the forms of A:B, exhibited in [A] and [B]. But if 
a?+6?=c?+d?; then (a*-+-b?)-(m?-+n?)=(c? +d?)-(m?+n?). 
Now (a? +-b?):(m? +-n? )=(amtbn)? + (anzbm)? =a? +62 =y?-+46? 
and (c*-+-d?)-(m? +-n?)= (emEdn)? +(enzdm)? = 4-22 = 7? +67, 
by employing the change of signs. But — 
(a? 4-62 )*(7? +62) = (a2 +6?) = (ay+55)?-++(a5z8y)?, and 
(«? +22)+(1 2 4.42)=(a2 +6)? =(entZa)? ++ (cbz%n)?. Thus we 
have found a square number, («?-+-¢?)?, which may be resolved into 
the sum of two squares in four different ways. In this way we may 
proceed, till we have found a square number that can be resolved 
into as many sets of squares as we please. But a better method is 
the following. 
