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x-2-4y-2-3z-2-x-y-z-N-HCF-x-y-z-1-like-4-1-2-x-y-z-lt-100-solve-by-computer-programing-




Question Number 176101 by BaliramKumar last updated on 12/Sep/22
x^2  − 4y^2  = 3z^2           [ x, y, z ∈ N ]             HCF(x, y, z) = 1                 like (4, 1, 2)                  x, y, z < 100  solve by computer programing
$${x}^{\mathrm{2}} \:−\:\mathrm{4}{y}^{\mathrm{2}} \:=\:\mathrm{3}{z}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\left[\:{x},\:{y},\:{z}\:\in\:\mathbb{N}\:\right]\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{HCF}\left({x},\:{y},\:{z}\right)\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{like}\:\left(\mathrm{4},\:\mathrm{1},\:\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x},\:{y},\:{z}\:<\:\mathrm{100} \\ $$$${solve}\:{by}\:{computer}\:{programing} \\ $$
Commented by Ar Brandon last updated on 12/Sep/22
#include <stdio.h> int main(void) { for (short x = 1; x < 100; x++) for (short y = 1; y < 100; y++) for (short z = 1; z < 100; z++) if (x*x - 4*y*y == 3*z*z) printf("(x,y,z)=(%hd,%hd,%hd)\n",x,y,z); return 0; }
Commented by Ar Brandon last updated on 12/Sep/22
(4, 1, 2), (14, 1, 8), (26, 11, 8), (28, 11, 10),  (28, 13, 6), (38, 13, 16), (52, 1, 30), (52, 23, 14),  (62, 23, 24), (74, 13, 40), (76, 11, 42), (76, 37, 10),  (86, 11, 48), (98, 47, 16).
$$\left(\mathrm{4},\:\mathrm{1},\:\mathrm{2}\right),\:\left(\mathrm{14},\:\mathrm{1},\:\mathrm{8}\right),\:\left(\mathrm{26},\:\mathrm{11},\:\mathrm{8}\right),\:\left(\mathrm{28},\:\mathrm{11},\:\mathrm{10}\right), \\ $$$$\left(\mathrm{28},\:\mathrm{13},\:\mathrm{6}\right),\:\left(\mathrm{38},\:\mathrm{13},\:\mathrm{16}\right),\:\left(\mathrm{52},\:\mathrm{1},\:\mathrm{30}\right),\:\left(\mathrm{52},\:\mathrm{23},\:\mathrm{14}\right), \\ $$$$\left(\mathrm{62},\:\mathrm{23},\:\mathrm{24}\right),\:\left(\mathrm{74},\:\mathrm{13},\:\mathrm{40}\right),\:\left(\mathrm{76},\:\mathrm{11},\:\mathrm{42}\right),\:\left(\mathrm{76},\:\mathrm{37},\:\mathrm{10}\right), \\ $$$$\left(\mathrm{86},\:\mathrm{11},\:\mathrm{48}\right),\:\left(\mathrm{98},\:\mathrm{47},\:\mathrm{16}\right). \\ $$
Commented by Ar Brandon last updated on 12/Sep/22
#include <iostream> using namespace std; int main(void) { for (short x = 1; x < 100; x++) for (short y = 1; y < 100; y++) for (short z = 1; z < 100; z++) if (x*x - 4*y*y == 3*z*z) cout<<"(x,y,z)=("<<x<<","<<y<<","<<z<<")"<<endl; return 0; }
Commented by BaliramKumar last updated on 13/Sep/22
thank you sir                 how taken time in programing  see ↓⇊⇊
$${thank}\:{you}\:{sir}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${how}\:{taken}\:{time}\:{in}\:{programing} \\ $$$${see}\:\downarrow\downdownarrows\downdownarrows \\ $$
Commented by BaliramKumar last updated on 12/Sep/22
Commented by BaliramKumar last updated on 13/Sep/22
A tringle′s  side in A.P. & area is rational  (a, b, c) = (3, 4, 5),  (13, 14, 15),  (15, 26, 37)                       (17, 28, 39), (15, 28, 41), (25, 38, 51)                      (51, 52, 53), (29, 52, 75), (39, 62, 85)                    ..............
$${A}\:{tringle}'{s}\:\:{side}\:{in}\:{A}.{P}.\:\&\:{area}\:{is}\:{rational} \\ $$$$\left({a},\:{b},\:{c}\right)\:=\:\left(\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right),\:\:\left(\mathrm{13},\:\mathrm{14},\:\mathrm{15}\right),\:\:\left(\mathrm{15},\:\mathrm{26},\:\mathrm{37}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{17},\:\mathrm{28},\:\mathrm{39}\right),\:\left(\mathrm{15},\:\mathrm{28},\:\mathrm{41}\right),\:\left(\mathrm{25},\:\mathrm{38},\:\mathrm{51}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{51},\:\mathrm{52},\:\mathrm{53}\right),\:\left(\mathrm{29},\:\mathrm{52},\:\mathrm{75}\right),\:\left(\mathrm{39},\:\mathrm{62},\:\mathrm{85}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………….. \\ $$

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