Search Results for ""
331 - 340 of 2395 for Burnside's theoremSearch Results
Euler's 6n+1 theorem states that every prime of the form 6n+1, (i.e., 7, 13, 19, 31, 37, 43, 61, 67, ..., which are also the primes of the form 3n+1; OEIS A002476) can be ...
The theorem, originally conjectured by Berge (1960, 1961), that a graph is perfect iff neither the graph nor its graph complement contains an odd graph cycle of length at ...
The asymptotic form of the n-step Bernoulli distribution with parameters p and q=1-p is given by P_n(k) = (n; k)p^kq^(n-k) (1) ∼ 1/(sqrt(2pinpq))e^(-(k-np)^2/(2npq)) (2) ...
Relates invariants of a curve defined over the integers. If this inequality were proven true, then Fermat's last theorem would follow for sufficiently large exponents. ...
Let a^p+b^p=c^p be a solution to Fermat's last theorem. Then the corresponding Frey curve is y^2=x(x-a^p)(x+b^p). (1) Ribet (1990a) showed that such curves cannot be modular, ...
The Feit-Thompson conjecture asserts that there are no primes p and q for which (p^q-1)/(p-1) and (q^p-1)/(q-1) have a common factor. Parker noticed that if this were true, ...
Specifying three sides uniquely determines a triangle whose area is given by Heron's formula, K=sqrt(s(s-a)(s-b)(s-c)), (1) where s=1/2(a+b+c) (2) is the semiperimeter of the ...
The point of coincidence of P and P^' in Fagnano's theorem.
Specifying three angles A, B, and C does not uniquely define a triangle, but any two triangles with the same angles are similar. Specifying two angles of a triangle ...
Specifying two angles A and B and a side a opposite A uniquely determines a triangle with area K = (a^2sinBsinC)/(2sinA) (1) = (a^2sinBsin(pi-A-B))/(2sinA). (2) The third ...
...
View search results from all Wolfram sites (5757 matches)

