Number Theory: An Introduction to Mathematics

1 Sums of Squares 543

Proposition 2The number of representations of a positive integer m as a sum of
2 squares of integers is equal to 4 times the excess of the number of positive divisors
of m of the form 4 h+ 1 over the number of positive divisors of the form 4 h+ 3.

Proof We h av e

θ 002 ( 0 )=

∑

n 1 ,n 2 ∈Z

qn

(^21) +n (^22)
= 1 +

∑

m≥ 1

r 2 (m)qm,

wherer 2 (m)is the number of solutions in integersn 1 ,n 2 of the equation

n^21 +n^22 =m.

To obtain another expression forθ^200 ( 0 )we use again the relation

θ 10 ′(v)/θ 10 (v)−θ 01 ′(v)/θ 01 (v)=πiθ 002 ( 0 )θ 00 (v)θ 11 (v)/θ 01 (v)θ 10 (v),

but this time we simply takev= 1 /4. Since

θ 01 ( 1 / 4 )=

∑

n∈Z

(−i)nqn

2

=

∑

n∈Z

i−nqn

2

=θ 00 ( 1 / 4 ),

and similarlyθ 11 ( 1 / 4 )=iθ 10 ( 1 / 4 ), we obtain

πθ 002 ( 0 )=θ 01 ′( 1 / 4 )/θ 01 ( 1 / 4 )−θ 10 ′( 1 / 4 )/θ 10 ( 1 / 4 ).

By differentiating logarithmically the product expansion forθ 10 (v)and then putting
v= 1 /4, we get

θ 10 ′( 1 / 4 )/θ 10 ( 1 / 4 )=−π− 4 π

∑

n≥ 1

q^2 n/( 1 +q^4 n).

Similarly, by differentiating logarithmically the product expansion forθ 01 (v)and then
puttingv= 1 /4, we get

θ 01 ′( 1 / 4 )/θ 01 ( 1 / 4 )= 4 π

∑

n≥ 1

q^2 n−^1 /( 1 +q^4 n−^2 ).

Thus

θ 01 ′( 1 / 4 )/θ 01 ( 1 / 4 )−θ 10 ′( 1 / 4 )/θ 10 ( 1 / 4 )=π+ 4 π

∑

n≥ 1

qn/( 1 +q^2 n)

and hence

θ 002 ( 0 )= 1 + 4

∑

n≥ 1

qn/( 1 +q^2 n).

Since

qn/( 1 +q^2 n)=qn( 1 −q^2 n)/( 1 −q^4 n)=(qn−q^3 n)

∑

k≥ 0

q^4 kn,