user55514
Reputation
454
Top tag
Next privilege 500 Rep.
Access review queues
 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 added 1 characters in body 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