a-2-b-2-c-2-a-b-c-1000-a-b-c-Z-Q-167533-reposted-

Question Number 167650 by Rasheed.Sindhi last updated on 22/Mar/22

Commented by Ari last updated on 22/Mar/22

Commented by mr W last updated on 22/Mar/22

i got only one solution for a,b,c∈N (200, 375, 425)

$${i}\:{got}\:{only}\:{one}\:{solution}\:{for}\:{a},{b},{c}\in\mathbb{N} \\ $$$$\left(\mathrm{200},\:\mathrm{375},\:\mathrm{425}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

But sir for a,b,c∈Z there are several solutions, I think.

$${But}\:\boldsymbol{{sir}}\:{for}\:{a},{b},{c}\in\mathbb{Z}\:{there}\:{are}\:{several} \\ $$$${solutions},\:{I}\:{think}. \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

$${a},{b},{c}\:{are}\:{integers}. \\ $$

Answered by Rasheed.Sindhi last updated on 22/Mar/22

determinant ((( { ((a=m^2 −n^2 )),((b=2mn)),((c=m^2 +n^2 )) :}∀ m,n∈Z}⇒a^2 +b^2 =c^2 ))) AND determinant (((a^2 +b^2 =c^2 ⇒ { ((a=m^2 −n^2 )),((b=2mn)),((c=m^2 +n^2 )) :} for some m,n∈Z))) a+b+c =(m^2 −n^2 )+(2mn)+(m^2 +n^2 )=1000 ⇒2m^2 +2mn=1000 m^2 +mn=500 n=((500−m^2 )/m)=((500)/m)−m ∵n∈Z ∴ m∣500 m=±1,±2,±4,±5,±10,±20,±25,±50,±100,±125,±250,±500 { ((a=m^2 −n^2 =m^2 −(((500)/m)−m)^2 )),((b=2mn=2m(((500)/m)−m))),((c=m^2 +n^2 =m^2 +(((500)/m)−m)^2 )) :} OR { ((a=2mn=2m(((500)/m)−m))),((b=m^2 −n^2 =m^2 −(((500)/m)−m)^2 )),((c=m^2 +n^2 =m^2 +(((500)/m)−m)^2 )) :} Where m is positive/negative divisor of 500. determinant ((m,n_(=((500)/m)−m) ,a_(=m^2 −n^2 _(=2mn) ) ,b_(=2mn_(=m^2 −n^2 ) ) ,c_(=m^2 +n^2 ) ),((±1),(±499),(−249000_(998) ),(998_(−249000) ),(249002_(249002) )),((±2),(±248),(−61500_(992) ),(992_(−61500) ),(61508_(61508) )),((±4),(±121),(−14625_(968) ),(968_(−14625) ),(14657_(14657) )),((±5),(±95),(−9000_(950) ),(950_(−9000) ),(9050_(9050) )),((±10),(±40),(−1500_(800) ),(800_(−1500) ),(1700_(1700) )),((±20),(±5),(375_(200) ),(200_(375) ),(425_(425) )),((±25),(∓5),(600_(−250) ),(−250_(600) ),(650_(650) )),((±50),(∓40),(900_(−4000) ),(−4000_(900) ),(4100_(4100) )),((±100),(∓95),(975_(−19000) ),(−19000_(975) ),(19025_(19025) )),((±125),(∓121),(984_(−30250) ),(−30250_(984) ),(30266_(30266) )),((±250),(∓248),(996_(−124000) ),(−124000_(996) ),(124004_(124004) )),((±500),(∓499),(999_(−499000) ),(−499000_(999) ),(499001_(499001) ))) 24 solutions

$$\begin{array}{|c|}{\left.\begin{cases}{{a}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} }\\{{b}=\mathrm{2}{mn}}\\{{c}={m}^{\mathrm{2}} +{n}^{\mathrm{2}} }\end{cases}\forall\:{m},{n}\in\mathbb{Z}\right\}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} }\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{AND} \\ $$$$\begin{array}{|c|}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \Rightarrow\begin{cases}{{a}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} }\\{{b}=\mathrm{2}{mn}}\\{{c}={m}^{\mathrm{2}} +{n}^{\mathrm{2}} }\end{cases}\:{for}\:{some}\:{m},{n}\in\mathbb{Z}}\\\hline\end{array} \\ $$$${a}+{b}+{c} \\ $$$$=\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)+\left(\mathrm{2}{mn}\right)+\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)=\mathrm{1000} \\ $$$$\Rightarrow\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{mn}=\mathrm{1000} \\ $$$$\:\:\:\:\:\:{m}^{\mathrm{2}} +{mn}=\mathrm{500} \\ $$$${n}=\frac{\mathrm{500}−{m}^{\mathrm{2}} }{{m}}=\frac{\mathrm{500}}{{m}}−{m} \\ $$$$\:\because{n}\in\mathbb{Z}\:\:\:\:\:\:\therefore\:{m}\mid\mathrm{500} \\ $$$${m}=\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{4},\pm\mathrm{5},\pm\mathrm{10},\pm\mathrm{20},\pm\mathrm{25},\pm\mathrm{50},\pm\mathrm{100},\pm\mathrm{125},\pm\mathrm{250},\pm\mathrm{500} \\ $$$$\begin{cases}{{a}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} ={m}^{\mathrm{2}} −\left(\frac{\mathrm{500}}{{m}}−{m}\right)^{\mathrm{2}} }\\{{b}=\mathrm{2}{mn}=\mathrm{2}{m}\left(\frac{\mathrm{500}}{{m}}−{m}\right)}\\{{c}={m}^{\mathrm{2}} +{n}^{\mathrm{2}} ={m}^{\mathrm{2}} +\left(\frac{\mathrm{500}}{{m}}−{m}\right)^{\mathrm{2}} }\end{cases} \\ $$$${OR} \\ $$$$\begin{cases}{{a}=\mathrm{2}{mn}=\mathrm{2}{m}\left(\frac{\mathrm{500}}{{m}}−{m}\right)}\\{{b}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} ={m}^{\mathrm{2}} −\left(\frac{\mathrm{500}}{{m}}−{m}\right)^{\mathrm{2}} }\\{{c}={m}^{\mathrm{2}} +{n}^{\mathrm{2}} ={m}^{\mathrm{2}} +\left(\frac{\mathrm{500}}{{m}}−{m}\right)^{\mathrm{2}} }\end{cases} \\ $$$${Where}\:{m}\:{is}\:{positive}/{negative} \\ $$$$\:{divisor}\:{of}\:\mathrm{500}. \\ $$$$ \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}{{m}}&\hline{\underset{=\frac{\mathrm{500}}{{m}}−{m}} {{n}}}&\hline{\underset{=\underset{=\mathrm{2}{mn}} {{m}^{\mathrm{2}} −{n}^{\mathrm{2}} }} {{a}}}&\hline{\underset{\underset{={m}^{\mathrm{2}} −{n}^{\mathrm{2}} } {=\mathrm{2}{mn}}} {{b}}}&\hline{\underset{={m}^{\mathrm{2}} +{n}^{\mathrm{2}} } {{c}}}\\{\pm\mathrm{1}}&\hline{\pm\mathrm{499}}&\hline{\underset{\mathrm{998}} {−\mathrm{249000}}}&\hline{\underset{−\mathrm{249000}} {\mathrm{998}}}&\hline{\underset{\mathrm{249002}} {\mathrm{249002}}}\\{\pm\mathrm{2}}&\hline{\pm\mathrm{248}}&\hline{\underset{\mathrm{992}} {−\mathrm{61500}}}&\hline{\underset{−\mathrm{61500}} {\mathrm{992}}}&\hline{\underset{\mathrm{61508}} {\mathrm{61508}}}\\{\pm\mathrm{4}}&\hline{\pm\mathrm{121}}&\hline{\underset{\mathrm{968}} {−\mathrm{14625}}}&\hline{\underset{−\mathrm{14625}} {\mathrm{968}}}&\hline{\underset{\mathrm{14657}} {\mathrm{14657}}}\\{\pm\mathrm{5}}&\hline{\pm\mathrm{95}}&\hline{\underset{\mathrm{950}} {−\mathrm{9000}}}&\hline{\underset{−\mathrm{9000}} {\mathrm{950}}}&\hline{\underset{\mathrm{9050}} {\mathrm{9050}}}\\{\pm\mathrm{10}}&\hline{\pm\mathrm{40}}&\hline{\underset{\mathrm{800}} {−\mathrm{1500}}}&\hline{\underset{−\mathrm{1500}} {\mathrm{800}}}&\hline{\underset{\mathrm{1700}} {\mathrm{1700}}}\\{\pm\mathrm{20}}&\hline{\pm\mathrm{5}}&\hline{\underset{\mathrm{200}} {\mathrm{375}}}&\hline{\underset{\mathrm{375}} {\mathrm{200}}}&\hline{\underset{\mathrm{425}} {\mathrm{425}}}\\{\pm\mathrm{25}}&\hline{\mp\mathrm{5}}&\hline{\underset{−\mathrm{250}} {\mathrm{600}}}&\hline{\underset{\mathrm{600}} {−\mathrm{250}}}&\hline{\underset{\mathrm{650}} {\mathrm{650}}}\\{\pm\mathrm{50}}&\hline{\mp\mathrm{40}}&\hline{\underset{−\mathrm{4000}} {\mathrm{900}}}&\hline{\underset{\mathrm{900}} {−\mathrm{4000}}}&\hline{\underset{\mathrm{4100}} {\mathrm{4100}}}\\{\pm\mathrm{100}}&\hline{\mp\mathrm{95}}&\hline{\underset{−\mathrm{19000}} {\mathrm{975}}}&\hline{\underset{\mathrm{975}} {−\mathrm{19000}}}&\hline{\underset{\mathrm{19025}} {\mathrm{19025}}}\\{\pm\mathrm{125}}&\hline{\mp\mathrm{121}}&\hline{\underset{−\mathrm{30250}} {\mathrm{984}}}&\hline{\underset{\mathrm{984}} {−\mathrm{30250}}}&\hline{\underset{\mathrm{30266}} {\mathrm{30266}}}\\{\pm\mathrm{250}}&\hline{\mp\mathrm{248}}&\hline{\underset{−\mathrm{124000}} {\mathrm{996}}}&\hline{−\underset{\mathrm{996}} {\mathrm{124000}}}&\hline{\underset{\mathrm{124004}} {\mathrm{124004}}}\\{\pm\mathrm{500}}&\hline{\mp\mathrm{499}}&\hline{\underset{−\mathrm{499000}} {\mathrm{999}}}&\hline{\underset{\mathrm{999}} {−\mathrm{499000}}}&\hline{\underset{\mathrm{499001}} {\mathrm{499001}}}\\\hline\end{array} \\ $$$$\mathrm{24}\:\mathrm{solutions} \\ $$

Commented by mr W last updated on 22/Mar/22

$${nice}\:{solution}! \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

$$\begin{array}{|c|}{\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\boldsymbol{\mathrm{sir}}!}\\\hline\end{array} \\ $$

Commented by Tawa11 last updated on 22/Mar/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Mar/22

$$\:\begin{array}{|c|}{\underset{\mathcal{M}{iss}} {\mathcal{TH}\alpha{nk}\mathcal{S}}}\\\hline\end{array} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Mar/22

Two solutions which are not covered by above approach are shared by my nephew Feroz (0,500,500) and (500,0,500).

$$\mathcal{T}{wo}\:{solutions}\:{which}\:{are}\:{not}\:{covered} \\ $$$${by}\:{above}\:{approach}\:{are}\:{shared}\:{by}\:{my} \\ $$$${nephew}\:{Feroz}\:\left(\mathrm{0},\mathrm{500},\mathrm{500}\right)\:{and} \\ $$$$\:\left(\mathrm{500},\mathrm{0},\mathrm{500}\right). \\ $$

Commented by Rasheed.Sindhi last updated on 29/Mar/22

{: (( a^2 +b^2 =c^2 )),((a+b+c=1000)) }; a^(?) , b^(?) , c^(?) ∈Z c^2 −b^2 =a^2 ⇒(c−b)(c+b)=a^2 A Case:c−b=c+b=a −b=b⇒b=0 { ((a^2 +b^2 =c^2 )),((a+b+c=1000)) :}⇒ { ((a^2 =c^2 )),((a+c=1000⇒a=1000−c)) :} a^2 =c^2 ⇒(1000−c)^2 =c^2 10^6 −2000c+c^2 =c^2 2000c=10^6 2c=1000 c=500 a=1000−c=1000−500=500 (a,b,c)=(500,0,500) Similarly (a,b,c)=(0,500,500)

$$\left.\begin{matrix}{\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} }\\{{a}+{b}+{c}=\mathrm{1000}}\end{matrix}\right\};\:\overset{?} {{a}},\:\:\overset{?} {{b}},\:\:\overset{?} {{c}}\in\mathbb{Z} \\ $$$$\:{c}^{\mathrm{2}} −{b}^{\mathrm{2}} ={a}^{\mathrm{2}} \Rightarrow\left({c}−{b}\right)\left({c}+{b}\right)={a}^{\mathrm{2}} \\ $$$$\mathrm{A}\:\mathrm{Case}:{c}−{b}={c}+{b}={a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−{b}={b}\Rightarrow{b}=\mathrm{0} \\ $$$$\begin{cases}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} }\\{{a}+{b}+{c}=\mathrm{1000}}\end{cases}\Rightarrow\begin{cases}{{a}^{\mathrm{2}} ={c}^{\mathrm{2}} }\\{{a}+{c}=\mathrm{1000}\Rightarrow{a}=\mathrm{1000}−{c}}\end{cases}\: \\ $$$${a}^{\mathrm{2}} ={c}^{\mathrm{2}} \Rightarrow\left(\mathrm{1000}−{c}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{10}^{\mathrm{6}} −\mathrm{2000}{c}+{c}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{2000}{c}=\mathrm{10}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{c}=\mathrm{1000} \\ $$$$\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{500} \\ $$$${a}=\mathrm{1000}−{c}=\mathrm{1000}−\mathrm{500}=\mathrm{500} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{500},\mathrm{0},\mathrm{500}\right) \\ $$$${Similarly}\:\left({a},{b},{c}\right)=\left(\mathrm{0},\mathrm{500},\mathrm{500}\right) \\ $$

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