MathLabs

Problem 1

Find all real numbers α\alpha so that, for every positive integer nn, the integer ⌊α⌋+⌊2α⌋+⌊3α⌋+⋯+⌊nα⌋\lfloor\alpha\rfloor+\lfloor2\alpha\rfloor+\lfloor3\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor is divisible by nn.
Step 2 of 4: Shift a non-integer into a short interval
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In plain words

Replacing alpha by alpha minus 2 changes the whole sum by a multiple of n(n+1), which is always a multiple of n, so the divisibility property is unaffected by shifting alpha by an even integer.

S(n,α−2)−S(n,α)=−n(n+1)≡0(modn)S(n,\alpha-2)-S(n,\alpha)=-n(n+1)\equiv0\pmod n
Detailed analysis

Suppose α\alpha is not an integer. Replacing α\alpha by α−2\alpha-2 changes each floor by exactly 22 times its index, so S(n,α−2)−S(n,α)=−2(1+2+⋯+n)=−n(n+1)≡0(modn)S(n,\alpha-2)-S(n,\alpha)=-2(1+2+\cdots+n)=-n(n+1)\equiv0\pmod n for every nn; hence the divisibility property is unchanged by shifting α\alpha by any even integer. So we may assume −1<α<1-1<\alpha<1 (and α≠0\alpha\ne0, since α=0\alpha=0 is an even integer already covered by Step 1).