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2h
comment Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
Seva: It doesn't seem to me that $p$ is assumed to be $1\pmod 8$. When $q$ divides $a$ (which is the example you raised) then the biquadratic symbol is determined by $(\frac{2}{q})$. I'm not an exper, and it may be safest to consult Lemmermeyer's book.
3h
comment Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
The sign doesn't matter. If you multiply $(\frac{\sigma(b+\sigma)}{q})$ and $(\frac{-\sigma(b-\sigma)}{q})$ together, you get $(\frac{-\sigma^2(b-\sigma^2)}{q}) = (\frac{a^2}{q})=1$.
4h
comment Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
An answer is given in the wikipedia page you linked in the section under Dirichlet. If $p\equiv \sigma^2 \pmod q$ then $(\frac{-q}{p})_4=(\frac{\sigma(b+\sigma)}{q})$. A reference is given to Lemmermeyer's book.
16h
reviewed Leave Closed Calculating the quotient group $\mathbb{Z}\times\mathbb{Z}/<(1,1),(1,-1)>$
1d
reviewed Leave Closed Robotics, Cryptography, and Genetics applications of Grothendieck's work?
1d
reviewed Close mathematical modelling question
1d
reviewed Close Solution to a system of linear equations containing some inequalities
1d
answered A perfect $(n,k)$ shuffle function
1d
comment A perfect $(n,k)$ shuffle function
If you label the cards $0$ to $n-1$ then your shuffle corresponds to multiplying card $i$ by the inverse of $k$ modulo $n-1$. Thus it returns to the original configuration in the order of $k^{-1}$ mod $n-1$ steps. But this is the same as the order of $k$ mod $n-1$.
1d
comment A perfect $(n,k)$ shuffle function
Seems to be the order of $k$ mod $n-1$. (This is well known for the usual perfect outer shuffle, and seems numerically to work in your other examples.)
2d
reviewed Close Find a prime when some primitive roots are given
Nov
20
reviewed Close “Almost” zeta function
Nov
20
reviewed Close elliptic curves and tower of finite fields
Nov
20
reviewed Close reduction of elliptic curves to finite field
Nov
20
reviewed Close Real number and axiom of continuity
Nov
20
reviewed No Action Needed Normal Covering of a Finite Group
Nov
19
reviewed Leave Open Finite extension of fields with no primitive element
Nov
19
reviewed Close Robotics, Cryptography, and Genetics applications of Grothendieck's work?
Nov
17
reviewed Leave Closed Is there a unique solution?
Nov
16
reviewed Leave Open Piecewise function of a pyramid surface