Equation with absolute values and parameters

Question

In my textbook there is a statement that for a fixed positive integer $a$, there is at most one positive integer $b$ such that: $|b-1|+|b-2|+\cdots+|b-2001|=a(a+1)$. They claim that it's obvious but to me it is not obvious. Can anyone explain why it's true?

Answer 1

The right hand side of the equation is equal to $\sum\limits_{k=1}^a 2k$ (from the formula for the sum of the first $a$ natural numbers. The left hand side, $|b-1|+|b-2|+\dots+|b-2001|$ can, if $0<b<2001$, be written as $\sum\limits_{k=1}^{b-1} k + \sum\limits_{k=1}^{2001-b} k$. Try to figure it out from here.

If $a=b=1001$ then the equality is true.

Answer 2

If $b \le 2001$, the summation

$\displaystyle \begin{align} \sum\limits_{k=1}^{2001}|b - k| = \sum\limits_{k=1}^{b}|b-k| + \sum\limits_{k=1+b}^{2001}|b-k| &= \frac{b^2-b}{2}+\frac{(2001-b)^2+(2001-b)}{2} \\ &= b^2 - 2002b + 2001\times 1001 \\ &= (b-1001)^2+1001000\end{align}$

The quadratic equation $a^2+a = (b-1001)^2+1000^2 + 1000$, one obvious solution is $b=1001$, then $a = 1000$.

For $a$ to be a positive integer the discriminant of the quadratic equation has to be perfect square.

That is, $1+4((b-1001)^2 + 1001000) = \Delta^2 \in \mathbb{Z}$.

$\implies 2001^2 + 4(b-1001)^2 = \Delta^2$

Recall that Pythageorean Triples are of the form: $(m^2-n^2)^2 + 4m^2n^2 = (m^2+n^2)^2$

Which means $(m-n)(m+n) = m^2-n^2 = 2001 = 3 \times 667 = 2001 \times 1$

Thus, the solutions for $(m,n) = (335,332)$ and $(1001,1000)$

Also, we must have $(b-1001)^2 = m^2n^2$, for either values of the pair $(m,n)$, the value of $mn > 1001$, which ensures unique solution of the pair $(a.b) = (1000,1001)$.

If $b > 2001$, the summation equals $2001b - 2001\times 1001 = 2001(b-1001) = a(a+1)$

$\implies (b-1001) \times 2001 = a(a+1)$

Where, the obvious solutions are $(a,b) = (2000,3001)$ and $(2001,3003)$.

So I guess the solution set is not unique after all.