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Math Help - Not the division algorithm (but fairly similar)

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    Not the division algorithm (but fairly similar)

    Prove that for any positive integer n, there exist unique positive integers k and r such that n = \frac{k(k-1)}{2}+r with 1\le r\le k.
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    Quote Originally Posted by Sigyn3 View Post
    Prove that for any positive integer n, there exist unique positive integers k and r such that n = \frac{k(k-1)}{2}+r with 1\le r\le k.
    Consider the set \{ n - \tfrac{x(x-1)}{2}\geq 0 | x\in \mathbb{Z}^+ \}. There exists a smallest integer n - \tfrac{k(k-1)}{2} = r. We claim that r\leq k. Assume not, then r>k and we can write r=r_0+k where r_0<r. Thus, we have n - \tfrac{k(k-1)}{2} = r_0+k which gives n-\tfrac{(k+1)k}{2} = r_0. And this is a contradiction since r was supposed to be minimal.
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    Does this also prove uniqueness?
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