Atrisinājums

Nosacījumu, ka trīs skaiţ̦i \(ab\), \(bc\), \(ac\) veido aritmētisko progresiju, pierakstīsim šādi:

\[ab + bc = 2ac, \text { jeb } a=\frac{bc}{2c-b}.\]

Apzīmēsim \((b, c)=d\); tad \(b=b_{1}d\), \(c=c_{1}d\), \(\left(b_{1}, c_{1}\right)=1\). No šejienes \(a=\frac{b_{1} c_{1} d}{2 b_{1}-c_{1}}\); tātad \(b_{1} c_{1} d\) dalās ar \(2 b_{1}-c_{1}\). Tā kā \(\left(b_{1}, 2 b_{1}-c_{1}\right) = \left(b_{1}, c_{1}\right) = 1\) un \(\left(c_{1}, 2b_{1} - c_{1} \right) = \left(2b_{1}, c_{1}\right) \in{1,2}\), tad pastāv divas iespējas: (a) \(\left(b_{1}, 2 b_{1}-c_{1}\right)=1,\left(c_{1}, 2 b_{1}-c_{1}\right)=1\); tad \(d=\left(2 b_{1}-c_{1}\right) k\). Iegūts sekojošs atrisinājums:

\[\begin{aligned} & b=\left(2 b_{1}-c_{1}\right) k, c=c_{1}\left(2 b_{1}-c_{1}\right) k, a=b_{1} c_{1} k \\ & b_{1}, c_{1}, k \in N, \quad\left(b_{1}, c_{1}\right)=1, \quad 2 b_{1}-c_{1}>0. \end{aligned}\]

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