hi voodoo child
st 1 can be proved with somewhat heavy math, but i am not sure that this way is optimal
(1)x^3+x=4k, where k is +ve integer.one thing to notice that sum of x^3+x results in some even integer that is 4k
we can obtain even integer from: even+even=even. or from odd+odd=even.
if x is even st 1 is suff, let us review what happens if x is odd
if x is odd, it must be perfomed as x=2m+1
here i`ii try to apply concept (a+b)^3=a^3+3a^2b+3ab^2+b^3
(2m+1)^3+(2m+1)=(8m^3+3*4m^2+3*2m+1)+2m+1=8m^3+3*4m^2+8m+2-but this expression is not divisible by 4 as 2 is not divisible by 4without remaider
thus if x is odd, this contradicts st1, and x must be even
