Question-6593

Question Number 6593 by Tawakalitu. last updated on 04/Jul/16

Commented by Yozzii last updated on 05/Jul/16

G i v e n t h a t p i s p r i m e a n d p ⩽ 100,

s h o w t h a t t h e r e a r e s o l u t i o n s (x, y),

x, y \in Z, f o r w h i c h y^{37} \equiv x^{3} + 37 (m o d p) .

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A n a t t e m p t \dots (p r o b a b l y m i s u n d e r s t o o d p r o b l e m)

\Rightarrow \exists r \in Z^{⩾} f o r p \in [2, 100] w i t h y^{37} - r p = x^{3} + 37

a n d y, x \in Z .

∴ r = \frac{y^{37} - x^{3} - 37}{p} . r ⩾ 0 ∴ y^{37} ⩾ x^{3} + 37

y^{37} ⩾ x^{3} + 27 + 10

y^{37} ⩾ (x + 3) (x^{2} - 3 x + 9) + 10 a n d t h e r e

a r e i n t e g e r s x, y s a t i s f y i n g t h i s i n e q u a l i t y .

r = \frac{y^{37} - (x + 3) (x^{2} - 3 x + 9) - 10}{p}

S u p p o s e p = 2 .

∴ r = \frac{y^{37} - (x^{3} + 1)}{2} - 18 .

I f x i s e v e n t h e n x^{3} + 1 i s o d d .

S o, f o r r \in Z^{⩾}, y^{37} m u s t b e o d d s o t h a t

y^{37} - x^{3} - 1 i s e v e n o r \exists k \in Z w h e r e

y^{37} - x^{3} - 1 = 2 k \Rightarrow r = k - 15 \in Z .

I f x i s o d d t h e n y i s e v e n . I n b o t h c a s e s

e v e n o r o d d i n t e g e r s x o r y c a n b e f o u n d a c c o r d i n g l y .

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A l l o t h e r p r i m e s a r e o d d a f t e r p = 2 .

r = \frac{y^{37} - (x^{3} + 1)}{p} - 18

c o n t i n u e \dots

Commented by prakash jain last updated on 05/Jul/16

y^{37} \equiv x^{3} + 37 (\mod p)

Since the question is a for limited set of

number i am just listing a solution .

case p ⩽ 37

p = 2 37 \equiv 1 \mod 2 \Rightarrow x = 0 y = 1

p = 3 37 \equiv 1 \mod 3 \Rightarrow x = 0 y = 1

p = 5 37 \equiv 2 \mod 5 \Rightarrow x = - 1 y = 1

p = 7 37 \equiv 2 \mod 7 \Rightarrow x = - 1 y = 1

p = 11 37 \equiv - 7 \mod 11 \Rightarrow x = 2 y = 1

p = 13 37 \equiv - 2 \mod 13 \Rightarrow x = 1 y = 1

p = 17 37 \equiv 3 \mod 17 \Rightarrow Solve y^{5} = x^{3} + 3 (m o d 17)

p = 19 37 \equiv - 1 \mod 19 \Rightarrow x = 1, y = 0

p = 23 37 \equiv - 9 \mod 23 \Rightarrow x = 2, y = - 1

p = 29 37 \equiv 8 \mod 29 \Rightarrow x = - 2, y = 0

p = 31 37 \equiv 6 \mod 31 \Rightarrow Solve y^{7} = x^{3} + 6 (m o d 31)

p \equiv 37 37 \equiv 0 \mod 37 \Rightarrow x = 0, y = 0