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Question Number 130085 by liberty last updated on 22/Jan/21
Show that if  { ((a=r^2 −2rs−s^2 )),((b=r^2 +s^2 )),((c=r^2 +2rs−s^2 )) :}   for some integers r,s then a^2 ,b^2 ,c^2   are three square in AP.
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\begin{cases}{\mathrm{a}=\mathrm{r}^{\mathrm{2}} −\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\\{\mathrm{b}=\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{c}=\mathrm{r}^{\mathrm{2}} +\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:\mathrm{r},\mathrm{s}\:\mathrm{then}\:\mathrm{a}^{\mathrm{2}} ,\mathrm{b}^{\mathrm{2}} ,\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{three}\:\mathrm{square}\:\mathrm{in}\:\mathrm{AP}. \\ $$
Answered by benjo_mathlover last updated on 22/Jan/21
 a^2 ,b^2 ,c^2  in AP then 2b^2  = a^2 +c^2   (1)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   (2) 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   It follows that a^2 ,b^2 ,c^2  in AP
$$\:{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{AP}\:\mathrm{then}\:\mathrm{2}{b}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\mathrm{2}{b}^{\mathrm{2}} =\mathrm{2}\left({r}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{2}\left({r}^{\mathrm{4}} +\mathrm{2}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +{s}^{\mathrm{4}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{r}^{\mathrm{4}} +\mathrm{4}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ac}\: \\ $$$$\:\:\:=\left(\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({r}^{\mathrm{2}} −{s}^{\mathrm{2}} −\mathrm{2}{rs}\right)\left({r}^{\mathrm{2}} −{s}^{\mathrm{2}} +\mathrm{2}{rs}\right) \\ $$$$\:=\mathrm{4}{r}^{\mathrm{4}} −\mathrm{8}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{4}{s}^{\mathrm{4}} −\mathrm{2}\left[\:\left({r}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} {s}^{\mathrm{2}} \:\right] \\ $$$$\:=\:\mathrm{4}{r}^{\mathrm{4}} −\mathrm{8}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{4}{s}^{\mathrm{4}} −\mathrm{2}\left({r}^{\mathrm{4}} −\mathrm{6}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +{s}^{\mathrm{4}} \right) \\ $$$$\:=\:\mathrm{2}{r}^{\mathrm{4}} +\mathrm{4}{r}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} \\ $$$$\mathrm{It}\:\mathrm{follows}\:\mathrm{that}\:{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{AP} \\ $$

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