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Question Number 130085 by liberty last updated on 22/Jan/21

$$\mathrm{Show}\mathrm{that}\mathrm{if}\{\begin{array}{l}\mathrm{a}={\mathrm{r}}^2-2\mathrm{rs}-{\mathrm{s}}^2\\ \mathrm{b}={\mathrm{r}}^2+{\mathrm{s}}^2\\ \mathrm{c}={\mathrm{r}}^2+2\mathrm{rs}-{\mathrm{s}}^2\end{array}$$$$\mathrm{for}\mathrm{some}\mathrm{integers}\mathrm{r},\mathrm{s}\mathrm{then}{\mathrm{a}}^2,{\mathrm{b}}^2,{\mathrm{c}}^2$$$$\mathrm{are}\mathrm{three}\mathrm{square}\mathrm{in}\mathrm{AP}.$$

Answered by benjo_mathlover last updated on 22/Jan/21

$$a^2,b^2,c^2\mathrm{in}\mathrm{AP}\mathrm{then}2b^2=a^2+c^2$$$$\left(1\right)2b^2=2{(r^2+s^2)}^2=2(r^4+2r^2{s}^2+s^4)$$$$=2r^4+4r^2{s}^2+2s^4$$$$\left(2\right)a^2+c^2={(a+c)}^2-2ac$$$$={(2r^2-2s^2)}^2-2(r^2-s^2-2rs)(r^2-s^2+2rs)$$$$=4r^4-8r^2{s}^2+4s^4-2[{(r^2-s^2)}^2-4r^2{s}^2]$$$$=4r^4-8r^2{s}^2+4s^4-2(r^4-6r^2{s}^2+s^4)$$$$=2r^4+4r^2{s}^2+2s^2$$$$\mathrm{It}\mathrm{follows}\mathrm{that}{a}^2,b^2,c^2\mathrm{in}\mathrm{AP}$$

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