*Posted by Billy in : potd , trackback
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Find the ordered triple of positive integers $(a,b,c)$ such that \[\begin{align*}a &= \frac{1908}{b(c-b)}-\frac{1908}{b(c+b)} \\ b &= \frac{2600}{a(c-a)}-\frac{2600}{a(c+a)} \\ c &= \frac{686}{a(b-a)}-\frac{686}{a(b+a)} .\end{align*}\]

show

We can rewrite each of the equations as \[\begin{align*} a\cdot c^2 - a\cdot b^2 &= 3816\\ b\cdot c^2 - b\cdot a^2 &= 5200 \\ c\cdot b^2 - c\cdot a^2 &= 1372.\end{align*}\] Subtracting the first and third equations from the second yields $$a\cdot b^2 – b\cdot a^2 + b\cdot c^2 – c\cdot b^2 + c\cdot a^2 – a\cdot c^2 = 12.$$ The left-hand side of this equation can be factored as $$(a-b)(b-c)(c-a).$$ It is easy to see that $a\neq b\neq c\neq 0$ and $(a-b) + (b-c) + (c-a) = 0$. Also, note that $c\cdot (b^2-a^2) = 1372$, and $c>0$, so $a<b$. Similarly, we find that $b<c$ and $a<c$, so $a<b<c$. As such, the only possibilities are that \[\begin{align*} -1 &= a-b \\ -3 &= b-c \\ 4 &= c-a \end{align*}\] or \[\begin{align*} -3 &= a-b \\ -1 &= b-c \\ 4 &= c-a. \end{align*}\] We now test the first case. Let $a=k$, $b=k+1$, and $c=k+4$. From the first equation of the problem statement, we have \[\begin{align*} 3816 &= k\cdot (k + 4)^2 - k\cdot (k + 1)^2 \\ &= 6k^2+15k .\end{align*}\] This quadratic equation yields $$k = \{24, -\frac{53}{2}\},$$ so $$(a,b,c) = \boxed{(24,25,28)}.$$ We do not need to check the other case because the problem statement implies that there is only one correct solution.

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