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

$a^{2} + b^{2} + c^{2} = 2019$ $a, b, c a r e p r i m e n u m b e r s .$ $h o w m a n y p o s s i b l e t r i p l e s o f (a, b, c) w h i c h t h a t s u i t a b l e f o r e q u a t i o n a b o v e .$
	Answered by afachri last updated on 01/Jan/19

	Commented by naka3546 last updated on 01/Jan/19

	$h o w t o g e t i t ?$
	Answered by mr W last updated on 02/Jan/19

	$s a y a ⩽ b ⩽ c$ $c ⩽ \sqrt{2019 - 2} \approx 45, i . e . t h e l a r g e s t p r i m e$ $c o u l d m a x . b e 43 .$ $\Rightarrow a, b, c \in (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43)$ $t h e s q u a r e s o f t h e s e n u m b e r s a r e$ $n n^{2}$ $1 1$ $2 4$ $3 9$ $5 25$ $7 49$ $11 121$ $13 169$ $17 289$ $19 361$ $23 529$ $29 841$ $31 961$ $37 1369$ $41 1681$ $43 1849$ $w e s t a r t w i t h c$ $c = 43 :$ $a^{2} + b^{2} = 2019 - 1849 = 170$ $\Rightarrow b = 13, a = 1 ∙$ $\Rightarrow b = 11, a = 7 ∙$ $c = 41 :$ $a^{2} + b^{2} = 2019 - 1681 = 338$ $b = 17 \Rightarrow a^{2} = 338 - 289 = 49 \Rightarrow a = 7 ∙$ $b = 13 \Rightarrow a^{2} = 338 - 13^{2} = 169 \Rightarrow a = 13 ∙$ $b = 11 \Rightarrow a^{2} = 338 - 11^{2} = 217 \Rightarrow n o s o l . f o r a ⩽ 11$ $\Rightarrow n o f u r t h e r s o l u t i o n f o r b < 11$ $c = 37 :$ $a^{2} + b^{2} = 2019 - 1369 = 650$ $b = 23 \Rightarrow a^{2} = 650 - 529 = 121 \Rightarrow a = 11 ∙$ $b = 19 \Rightarrow a^{2} = 650 - 361 = 289 \Rightarrow a = 17 ∙$ $b = 17 \Rightarrow a^{2} = 650 - 289 = 361 \Rightarrow n o s o l . f o r a ⩽ 17$ $\Rightarrow n o f u r t h e r s o l u t i o n f o r b < 17$ $c = 31 :$ $a^{2} + b^{2} = 2019 - 961 = 1058$ $b = 31 \Rightarrow a^{2} = 1058 - 961 = 97 \Rightarrow n o s o l . f o r a ⩽ 31$ $b = 29 \Rightarrow a^{2} = 1058 - 841 = 217 \Rightarrow n o s o l . f o r a ⩽ 29$ $b = 23 \Rightarrow a^{2} = 1058 - 529 = 529 \Rightarrow a = 23 ∙$ $\Rightarrow n o f u r t h e r s o l u t i o n f o r b < 23$ $c = 29 :$ $a^{2} + b^{2} = 2019 - 841 = 1178$ $b = 29 \Rightarrow a^{2} = 1178 - 841 = 337 \Rightarrow n o s o l . f o r a ⩽ 29$ $b = 23 \Rightarrow a^{2} = 1178 - 529 = 649 \Rightarrow n o s o l . f o r a ⩽ 23$ $\Rightarrow n o f u r t h e r s o l u t i o n f o r b < 23$ $\Rightarrow n o f u r t h e r s o l u t i o n f o r c < 29$ $\Rightarrow a l l s o l u t i o n s a r e$ $(1, 13, 43)$ $(7, 11, 43)$ $(7, 17, 41)$ $(13, 13, 41)$ $(11, 23, 37)$ $(17, 19, 37)$ $(23, 23, 31)$
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