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 4 of 4: Case -1 < alpha < 0 fails, concluding alpha is an even integer
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In plain words

For small negative alpha, every floor equals -1 until the running sum finally drops below -1, at which point it becomes -2, giving a total that is one short of a multiple of m and so is never divisible by m.

−1<α<0  ⟹  S(m,α)=−(m+1)-1<\alpha<0\implies S(m,\alpha)=-(m+1)
Detailed analysis

If −1<α<0-1<\alpha<0, let m≥2m\ge2 be the smallest integer with mα≤−1m\alpha\le-1; then ⌊iα⌋=−1\lfloor i\alpha\rfloor=-1 for i=1,…,m−1i=1,\ldots,m-1 and ⌊mα⌋=−2\lfloor m\alpha\rfloor=-2 (by minimality mα>−2m\alpha>-2), giving S(m,α)=−(m−1)−2=−(m+1)S(m,\alpha)=-(m-1)-2=-(m+1), which is not divisible by mm since gcd⁡(m,m+1)=1\gcd(m,m+1)=1 and m≥2m\ge2. Both sub-cases of a non-integer α\alpha lead to a contradiction, so combined with Step 1, α\alpha must be an even integer.