find all integers n for which 13 ∣4(n^2 +1).


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Question Number 94097 by Rio Michael last updated on 16/May/20

$find all integers n for which 13 ∣ 4 (n^{2} + 1) .$
Commented by Rasheed.Sindhi last updated on 17/May/20

$T h i s s e q u i v a l e n t t o :$ $13 ∣ (n^{2} + 1)$ $\frac{n^{2} + 1}{13} = k \Rightarrow 13 k - 1 = n^{2}$ $A l l k^{'} s f o r w h i c h 13 k - 1 i s$ $p e r f e c t s q u a r e .$ $S o m e v a l u e s o f k :$ $k = 2, 5, 25, 34, 74, 89, . . .$ $n = 5, 8, 18, 21, 31, 34, 44, . . .$ $G e n e r a l s o l u t i o n s :$ $n = 13 q + 5, 13 q + 8 \forall q \in Z$ $. . . . .$ $. . .$
	Answered by Rasheed.Sindhi last updated on 17/May/20
	$13 ∤ 4⇒13 ∣ (n^2 +1) Let n=13k+i where k,i∈Z ∧ 0≤i≤12 n^2 +1=13^2 k^2 +2(13k)(i)+i^2 +1 13 ∣ (n^2 +1)⇒13∣{13^2 k^2 +2(13k)(i)+i^2 +1} ⇒ 13 ∣ (i^2 +1) In interval 0≤i≤12 only i=5,8 satisfy above. Hence n=13k+5,13k+8 are satisfying values ∀ k∈Z$
	$13 ∤ 4 \Rightarrow 13 ∣ (n^{2} + 1)$ $L e t n = 13 k + i w h e r e k, i \in Z \land 0 ⩽ i ⩽ 12$ $n^{2} + 1 = 13^{2} k^{2} + 2 (13 k) (i) + i^{2} + 1$ $13 ∣ (n^{2} + 1) \Rightarrow 13 ∣ {13^{2} k^{2} + 2 (13 k) (i) + i^{2} + 1}$ $\Rightarrow 13 ∣ (i^{2} + 1)$ $I n i n t e r v a l 0 ⩽ i ⩽ 12 o n l y i = 5, 8$ $s a t i s f y a b o v e .$ $H e n c e$ $n = 13 k + 5, 13 k + 8 a r e s a t i s f y i n g$ $v a l u e s \forall k \in Z$
	Answered by Rasheed.Sindhi last updated on 17/May/20

	$13 ∣ 4 (n^{2} + 1) i s a q u i v a l e n t t o :$ $13 ∣ (n^{2} + 1)$ $^{∙} n : 13 k, 13 k + 1, . . ., 13 k + 4, 13 k + 5,$ $13 k + 6, 13 k + 7, 13 k + 8, 13 k + 9,$ $. . ., 13 k + 12$ $^{∙} F o r v a l u e s o f n w r i t t e n i n b l a c k$ $W e c a n e a s i l y p r o v e t h a t :$ $13 ∤ (n^{2} + 1)$ $F o r e x a m p l e i f n = 13 k + 2$ $n^{2} + 1 = 13^{2} k^{2} + 2.13 k . 2 + 5$ $= 13 (13 k^{2} + 4 k) + 5$ $∴ 13 ∤ (n^{2} + 1)$ $^{∙} F o r v a l u e s o f n w r i t t e n i n r e d$ $W e c a n e a s i l y p r o v e t h a t :$ $13 ∣ (n^{2} + 1)$ $F o r a n e x a m p l e i f n = 13 k + 8$ $n^{2} + 1 = 13^{2} k^{2} + 2.13 k . 8 + 65$ $= 13 (13 k^{2} + 16 k + 5)$ $∴ 13 ∣ (n^{2} + 1)$ $^{∙} H e n c e 13 ∣ (n^{2} + 1) f o r o n l y$ $13 k + 5 & 13 k + 8$
	Commented by Rio Michael last updated on 17/May/20

	$thank you sir .$
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