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 1 of 4: Integers: even works, odd fails
Previous stepNext step
In plain words

When alpha is an integer, every floor vanishes and the sum becomes a simple closed form, which is divisible by n for every n exactly when alpha is even.

S(n,α)=α⋅n(n+1)2S(n,\alpha)=\alpha\cdot\dfrac{n(n+1)}2
Detailed analysis

If α\alpha is an integer, S(n,α):=⌊α⌋+⋯+⌊nα⌋=α(1+2+⋯+n)=α⋅n(n+1)2S(n,\alpha):=\lfloor\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor=\alpha(1+2+\cdots+n)=\alpha\cdot\dfrac{n(n+1)}2. Requiring n∣S(n,α)n\mid S(n,\alpha) for all nn means α(n+1)2∈Z\dfrac{\alpha(n+1)}2\in\mathbb Z for all nn; this holds for every nn exactly when α\alpha is even. If α\alpha is an odd integer, n=2n=2 gives S(2,α)=3αS(2,\alpha)=3\alpha, which is odd and not divisible by 22, so odd integers fail.