( 434 ) 
One square must be 
1+34+54+....+t4a, 
the other 
58 aPbO re Ene ey 
So we have: 
G= 3024 53 
257-50 fa = 6D 
292 1 112 
57 
28? 4 169 
55 
2 
Zok So G = 28° +137. 
dd 
51 
49 farther when half of G is attained. 
47 
424 
It is unnecessary to continue 
A square is always a 4-fold or a 4-fold + 1, so the sum of two 
squares is a 4-fold or a 4-fold +1 or a 4-fold + 2. 
So a 4-fold —1 can never be decomposed into the sum of two 
squares. 
As a special case of the property sub. 1 we have: 
2. Every indivisible 4-fold + 1 can be decomposed in one way 
into the sum of two squares. 
If a 4-fold + 1 contains factors which are 4-fold —1, the 
decomposition is not always possible. 
Example: G = 957=3 X11 = 29. The factors 3 and 11 are 
an objection to the decomposition. 
From this follows immediately : 
3. If a 4-fold +1 cannot be decomposed into the sum of two 
squares, il is divisible. 
