In number theory and harmonic analysis, the Landsberg–Schaar relation (or identity) is the following equation, which is valid for arbitrary positive integers p and q:
1 p ∑ n = 0 p − 1 exp ( 2 π i n 2 q p ) = e 1 4 π i 2 q ∑ n = 0 2 q − 1 exp ( − π i n 2 p 2 q ) . {\displaystyle {\frac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\frac {e^{{\frac {1}{4}}\pi i}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right).}
The standard way to prove it is to put τ = 2iq/p + ε, where ε > 0 in this identity due to Jacobi (which is essentially just a special case of the Poisson summation formula in classical harmonic analysis):
∑ n = − ∞ + ∞ e − π n 2 τ = 1 τ ∑ n = − ∞ + ∞ e − π n 2 τ {\displaystyle \sum _{n=-\infty }^{+\infty }e^{-\pi n^{2}\tau }={\frac {1}{\sqrt {\tau }}}\sum _{n=-\infty }^{+\infty }e^{-\pi {\frac {n^{2}}{\tau }}}}
and then let ε → 0. A proof using only finite methods was discovered in 2018 by Ben Moore. If we let q = 1, the identity reduces to a formula for the quadratic Gauss sum modulo p. The Landsberg–Schaar identity can be rephrased more symmetrically as
1 p ∑ n = 0 p − 1 exp ( π i n 2 q p ) = e 1 4 π i q ∑ n = 0 q − 1 exp ( − π i n 2 p q ) {\displaystyle {\frac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {\pi in^{2}q}{p}}\right)={\frac {e^{{\frac {1}{4}}\pi i}}{\sqrt {q}}}\sum _{n=0}^{q-1}\exp \left(-{\frac {\pi in^{2}p}{q}}\right)}
provided that we add the hypothesis that pq is an even number.
References
