problemes de enters

sum[k=0-->p]( (q/p)·k ) + (-1)·sum[k=0-->q]( (p/q)·k ) = (1/2)·(q+(-p))


sum[k=0-->p]( (q/p)(q+1)·k^{2} ) + (-1)·sum[k=0-->q]( (p/q)(p+1)·k^{2} ) = ...
...(1/6)·(q+(-p))(p+1)(q+1)


sum[k=0-->p]( (q/p^{2})(q+1)·k^{3} ) + (-1)·sum[k=0-->q]( (p/q^{2})(p+1)·k^{3} ) = ...
...(1/4)·(q+(-p))(p+1)(q+1)




proposició:
si n=2^{k}+(-1) ==> [Ep][ p€N & 2n+1=2^{p}+(-1) ]
demostració
2n+1=2·(2^{k}+(-1))+1=2^{k+1}+(-2)+1
2n+1=2^{k+1}+(-1)=2^{p}+(-1)


proposició:
si n=m^{k}+(-1) ==> [Ep][ p€N & m·n+(m+(-1))=m^{p}+(-1) ]
proposició:
si n=m^{k}+1 ==> [Ep][ p€N & m·n+(-1)(m+(-1))=m^{p}+1 ]


a=mb_{a}+r_{a}
a+(-m)b_{a}=r_{a}


sum[a]( (a+(-m)b_{a})·(1/m(m+(-1))) )=(1/2)


sum[a]( (a+(-m)b_{a})^{2}·(1/m(m+(-1))(2m+(-1))) )=(1/6)


sum[a]( (a+(-m)b_{a})^{3}·(1/m^{2}(m+(-1))^{2}) )=(1/4)