find-all-integers-n-for-which-13-4-n-2-1-

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, \dots

n = 5, 8, 18, 21, 31, 34, 44, \dots

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

\dots . .

\dots

Answered by Rasheed.Sindhi last updated on 17/May/20

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, \dots, 13 k + 4, 13 k + 5,

13 k + 6, 13 k + 7, 13 k + 8, 13 k + 9,

\dots, 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 .