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Nov
4
awarded  Nice Question
Nov
4
comment How to measure the irregularity of a hexagon?
Very interesting problem. No time to try to answer it now, though i thought very briefly about it, once. Here is something in Spanish language. How many hexagons you can find with equal internal angles and sides of consecutive integral lengths. lit-et-raire.blogspot.com.es/2012/08/poligonos-y-enteros.html
Oct
28
answered Prove that all values $n$ which are odd can be written in one of two forms: $4q+1$, $4q+3$ with a non-negative integer $q$?
Oct
19
awarded  Nice Answer
Oct
5
awarded  Yearling
Jul
2
awarded  Curious
Jan
7
awarded  Yearling
Sep
23
awarded  Necromancer
Feb
26
comment Cumulative probabilities
I did not; thanks.
Feb
26
revised Cumulative probabilities
added 35 characters in body
Feb
26
revised Cumulative probabilities
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Feb
26
asked Cumulative probabilities
Feb
19
comment Least power. Squares again
Ok; i see know. But it would have been clearer to say Product () -Product () = 2f(0) = constant. I thought, first, c was an arbitray external constant. This is why i did not iunderstand.
Feb
19
comment Least power. Squares again
Though there is the short posibility of g(0)=2(2k+1)=4k+2 with no solution.
Feb
19
comment Least power. Squares again
I do not understand quite well the step where it is said that Product(x+ai)-Product(x+bi) is a constant
Feb
19
comment Least power. Squares again
Andreas, your proof seems correct to me. Just you should not say g(0) is odd "avant la lettre". In my opinion g(0) must be odd because u and v have distint parity and we must have uv = 2^(k/2-1)g(0)^2 so g(0) must be odd. Also i agree k must be a multiple of 4, because half the multiplied terms are negative and a square is always positive. The only small doubt i have is the validity/generality of the choosen x= (k-1)/2, a non integer value for x; to give the full proof.
Feb
18
revised Least power. Squares again
deleted 48 characters in body
Feb
18
comment Least power. Squares again
Thank you for the answers. I thought i had a proof by induction for the case k=4 but it was not a correct proof. Sorry for it.
Feb
17
revised Least power. Squares again
added 1 characters in body
Feb
17
asked Least power. Squares again