How many ways can 2018 be expressed as the sum of two squares?


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Question Number 110895 by Aina Samuel Temidayo last updated on 31/Aug/20

$How many ways can 2018 be$ $expressed as the sum of two squares ?$
	Answered by ajfour last updated on 31/Aug/20

	$43^{2} + 13^{2} = 2018$
	Answered by floor(10²Eta[1]) last updated on 31/Aug/20
	$y=(√(2018−x^2 ))⇒x<45 and y<45 x^2 +y^2 =2018⇒(x+y)^2 =2(1009+xy) (x+y)^2 is even ⇒(x+y) is even. suppose x and y are even x=2a and y=2b (2a+2b)^2 =2(1009+(2a)(2b)) ⇒2(a^2 +b^2 )=1009 but 1009 is not even so x and y can′t be even. ⇒ x and y are odd. gcd(x,y)=d⇒x=ds, y=dt, gcd(s, t)=1 d^2 (s^2 +t^2 )=2018 d^2 ∣2018⇒d^2 ∈{1, 2, 1009, 2018}⇒d=1 so x and y are coprimes x,y∈(1,3,5,7,...,43) with that you can try to solve the problem by testing but i will continue to think and try to get a better way$
	$y = \sqrt{2018 - x^{2}} \Rightarrow x < 45 and y < 45$ $x^{2} + y^{2} = 2018 \Rightarrow {(x + y)}^{2} = 2 (1009 + xy)$ ${(x + y)}^{2} is even \Rightarrow (x + y) is even .$ $suppose x and y are even$ $x = 2 a and y = 2 b$ ${(2 a + 2 b)}^{2} = 2 (1009 + (2 a) (2 b))$ $\Rightarrow 2 (a^{2} + b^{2}) = 1009$ $but 1009 is not even so x and y {can}^{'} t be even .$ $\Rightarrow x and y are odd .$ $\gcd (x, y) = d \Rightarrow x = ds, y = dt, \gcd (s, t) = 1$ $d^{2} (s^{2} + t^{2}) = 2018$ $d^{2} ∣ 2018 \Rightarrow d^{2} \in {1, 2, 1009, 2018} \Rightarrow d = 1$ $so x and y are coprimes$ $x, y \in (1, 3, 5, 7, . . ., 43)$ $with that you can try to solve the problem$ $by testing but i will continue to think$ $and try to get a better way$
	Commented by Aina Samuel Temidayo last updated on 31/Aug/20

	$Thanks$
	Answered by Rasheed.Sindhi last updated on 01/Sep/20

	$(i) unit - digits .$ $(ii) squares of unit - digits .$ $(iii) unit - digits of squares$ $[\begin{matrix} (i) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ (ii) & 0 & 1 & 4 & 9 & 16 & 25 & 36 & 49 & 64 & 81 \\ (iii) & 0 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \end{matrix}]$ $i s s u m o f t w o s q u a r e n u m b e r s .$ $H e n c e i t s u n i t - d i g i t (8) i s m a d e u p$ $o f s u m o f u n i t - d i g i t s o f s q u a r e s$ $O n l y t w o p o s s i b i l i t i e s :$ $4 + 4 = 8 & 9 + 9 = 18$ $^{∙} 4 i s s q u a r e o f u n i t - d i g i t s : 2 & 8$ $^{∙} 9 i s s q u a r e o f u n i t - d i g i t s : 3 & 7$ $S e a r c h f o r t h e n u m b e r s i s$ $n a r r o w e n o u g h n o w : O n l y$ $n u m b e r s w i t h u n i t s 2, 8, 3 & 7$ $S o f i n a l l y 3 i s s u c c e s s f u l$ $c a n d i d a t e . I f n i s o n e p o s s i b l e$ $c a n d i d a t e f o r r e q u i r e d n u m b e r s$ $o t h e r n u m b e r N i s$ $N = \sqrt{2018 - n}$ $p r o v i d e d t h a t 2018 - n i s p r e f e c t$ $s q u r e .$ $F i n a l l y 13^{2} + 43^{2} = 2018 o n l y$ $s o l u t i o n .$
	Commented by floor(10²Eta[1]) last updated on 01/Sep/20

	$cool!! with that and knowing that$ $the numbers are both odd helps a lot$ $because we just have to look for$ $the numbers (3, 7, 13, 17, 23, 27, 33, 37, 43)$
	Commented by Rasheed.Sindhi last updated on 01/Sep/20

	$T han k s floor sir!$
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