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a-2-b-2-c-2-2019-a-b-c-are-prime-numbers-how-many-possible-triples-of-a-b-c-which-that-suitable-for-equation-above-

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Question Number 51950 by naka3546 last updated on 01/Jan/19
a^2 + b^2 + c^2 = 2019 a, b, c are prime numbers . how many possible triples of (a, b, c) which that suitable for equation above .
$${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:\:=\:\:\mathrm{2019} \\ $$$${a},\:{b},\:{c}\:\:\:{are}\:\:{prime}\:\:{numbers}\:. \\ $$$${how}\:\:{many}\:\:{possible}\:\:{triples}\:\:{of}\:\:\left({a},\:{b},\:{c}\right)\:\:{which}\:\:{that}\:\:{suitable}\:\:{for}\:\:{equation}\:\:{above}\:. \\ $$
Answered by afachri last updated on 01/Jan/19
Commented by naka3546 last updated on 01/Jan/19
how to get it ?
$${how}\:\:{to}\:\:{get}\:\:{it}\:\:? \\ $$
Answered by mr W last updated on 02/Jan/19
say a≤b≤c c≤(√(2019−2))≈45, i.e. the largest prime could max. be 43. ⇒a,b,c ∈(2,3,5,7,11,13,17,19,23,29,31,37,41,43) the squares of these numbers are n n^2 1 1 2 4 3 9 5 25 7 49 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 1369 41 1681 43 1849 we start with c c=43: a^2 +b^2 =2019−1849=170 ⇒b=13, a=1 • ⇒b=11, a=7 • c=41: a^2 +b^2 =2019−1681=338 b=17⇒a^2 =338−289=49⇒a=7 • b=13⇒a^2 =338−13^2 =169⇒a=13 • b=11⇒a^2 =338−11^2 =217⇒no sol. for a≤11 ⇒no further solution for b<11 c=37: a^2 +b^2 =2019−1369=650 b=23⇒a^2 =650−529=121⇒a=11 • b=19⇒a^2 =650−361=289⇒a=17 • b=17⇒a^2 =650−289=361⇒no sol. for a≤17 ⇒no further solution for b<17 c=31: a^2 +b^2 =2019−961=1058 b=31⇒a^2 =1058−961=97⇒no sol. for a≤31 b=29⇒a^2 =1058−841=217⇒no sol. for a≤29 b=23⇒a^2 =1058−529=529⇒a=23 • ⇒no further solution for b<23 c=29: a^2 +b^2 =2019−841=1178 b=29⇒a^2 =1178−841=337⇒no sol. for a≤29 b=23⇒a^2 =1178−529=649⇒no sol. for a≤23 ⇒no further solution for b<23 ⇒no further solution for c<29 ⇒all solutions are (1, 13, 43) (7, 11, 43) (7, 17, 41) (13, 13, 41) (11, 23, 37) (17, 19, 37) (23, 23, 31)
$${say}\:{a}\leqslant{b}\leqslant{c} \\ $$$${c}\leqslant\sqrt{\mathrm{2019}−\mathrm{2}}\approx\mathrm{45},\:{i}.{e}.\:{the}\:{largest}\:{prime} \\ $$$${could}\:{max}.\:{be}\:\mathrm{43}. \\ $$$$\Rightarrow{a},{b},{c}\:\in\left(\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{11},\mathrm{13},\mathrm{17},\mathrm{19},\mathrm{23},\mathrm{29},\mathrm{31},\mathrm{37},\mathrm{41},\mathrm{43}\right) \\ $$$${the}\:{squares}\:{of}\:{these}\:{numbers}\:{are} \\ $$$${n}\:\:\:\:\:\:\:{n}^{\mathrm{2}} \\ $$$$\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{9} \\ $$$$\mathrm{5}\:\:\:\:\:\:\:\:\mathrm{25} \\ $$$$\mathrm{7}\:\:\:\:\:\:\:\:\mathrm{49} \\ $$$$\mathrm{11}\:\:\:\:\:\mathrm{121} \\ $$$$\mathrm{13}\:\:\:\:\:\mathrm{169} \\ $$$$\mathrm{17}\:\:\:\:\:\mathrm{289} \\ $$$$\mathrm{19}\:\:\:\:\:\mathrm{361} \\ $$$$\mathrm{23}\:\:\:\:\:\mathrm{529} \\ $$$$\mathrm{29}\:\:\:\:\:\mathrm{841} \\ $$$$\mathrm{31}\:\:\:\:\:\mathrm{961} \\ $$$$\mathrm{37}\:\:\:\:\:\mathrm{1369} \\ $$$$\mathrm{41}\:\:\:\:\:\mathrm{1681} \\ $$$$\mathrm{43}\:\:\:\:\:\mathrm{1849} \\ $$$$ \\ $$$${we}\:{start}\:{with}\:{c} \\ $$$${c}=\mathrm{43}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}−\mathrm{1849}=\mathrm{170} \\ $$$$\Rightarrow{b}=\mathrm{13},\:{a}=\mathrm{1}\:\:\bullet\:\: \\ $$$$\Rightarrow{b}=\mathrm{11},\:{a}=\mathrm{7}\:\:\bullet \\ $$$$ \\ $$$${c}=\mathrm{41}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}−\mathrm{1681}=\mathrm{338} \\ $$$${b}=\mathrm{17}\Rightarrow{a}^{\mathrm{2}} =\mathrm{338}−\mathrm{289}=\mathrm{49}\Rightarrow{a}=\mathrm{7}\:\:\bullet \\ $$$${b}=\mathrm{13}\Rightarrow{a}^{\mathrm{2}} =\mathrm{338}−\mathrm{13}^{\mathrm{2}} =\mathrm{169}\Rightarrow{a}=\mathrm{13}\:\:\bullet \\ $$$${b}=\mathrm{11}\Rightarrow{a}^{\mathrm{2}} =\mathrm{338}−\mathrm{11}^{\mathrm{2}} =\mathrm{217}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{11} \\ $$$$\Rightarrow{no}\:{further}\:{solution}\:{for}\:{b}<\mathrm{11} \\ $$$$ \\ $$$${c}=\mathrm{37}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}−\mathrm{1369}=\mathrm{650} \\ $$$${b}=\mathrm{23}\Rightarrow{a}^{\mathrm{2}} =\mathrm{650}−\mathrm{529}=\mathrm{121}\Rightarrow{a}=\mathrm{11}\:\:\bullet \\ $$$${b}=\mathrm{19}\Rightarrow{a}^{\mathrm{2}} =\mathrm{650}−\mathrm{361}=\mathrm{289}\Rightarrow{a}=\mathrm{17}\:\:\bullet \\ $$$${b}=\mathrm{17}\Rightarrow{a}^{\mathrm{2}} =\mathrm{650}−\mathrm{289}=\mathrm{361}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{17} \\ $$$$\Rightarrow{no}\:{further}\:{solution}\:{for}\:{b}<\mathrm{17} \\ $$$$ \\ $$$${c}=\mathrm{31}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}−\mathrm{961}=\mathrm{1058} \\ $$$${b}=\mathrm{31}\Rightarrow{a}^{\mathrm{2}} =\mathrm{1058}−\mathrm{961}=\mathrm{97}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{31} \\ $$$${b}=\mathrm{29}\Rightarrow{a}^{\mathrm{2}} =\mathrm{1058}−\mathrm{841}=\mathrm{217}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{29} \\ $$$${b}=\mathrm{23}\Rightarrow{a}^{\mathrm{2}} =\mathrm{1058}−\mathrm{529}=\mathrm{529}\Rightarrow{a}=\mathrm{23}\:\:\bullet \\ $$$$\Rightarrow{no}\:{further}\:{solution}\:{for}\:{b}<\mathrm{23} \\ $$$$ \\ $$$${c}=\mathrm{29}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}−\mathrm{841}=\mathrm{1178} \\ $$$${b}=\mathrm{29}\Rightarrow{a}^{\mathrm{2}} =\mathrm{1178}−\mathrm{841}=\mathrm{337}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{29} \\ $$$${b}=\mathrm{23}\Rightarrow{a}^{\mathrm{2}} =\mathrm{1178}−\mathrm{529}=\mathrm{649}\Rightarrow{no}\:{sol}.\:{for}\:{a}\leqslant\mathrm{23} \\ $$$$\Rightarrow{no}\:{further}\:{solution}\:{for}\:{b}<\mathrm{23} \\ $$$$ \\ $$$$\Rightarrow{no}\:{further}\:{solution}\:{for}\:{c}<\mathrm{29} \\ $$$$ \\ $$$$\Rightarrow{all}\:{solutions}\:{are}\: \\ $$$$\left(\mathrm{1},\:\mathrm{13},\:\mathrm{43}\right) \\ $$$$\left(\mathrm{7},\:\mathrm{11},\:\mathrm{43}\right) \\ $$$$\left(\mathrm{7},\:\mathrm{17},\:\mathrm{41}\right) \\ $$$$\left(\mathrm{13},\:\mathrm{13},\:\mathrm{41}\right) \\ $$$$\left(\mathrm{11},\:\mathrm{23},\:\mathrm{37}\right) \\ $$$$\left(\mathrm{17},\:\mathrm{19},\:\mathrm{37}\right) \\ $$$$\left(\mathrm{23},\:\mathrm{23},\:\mathrm{31}\right) \\ $$

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