Question

(2014k + 1)(2018k + 1) is a perfect square number. find the smallest value of k with k is natural number

Answer 1

can anyone here help me in my ques

Answer 2

I don't see how this is possible unless 1 is a perfect square.

Answer 3

So k=0 I guess?

Answer 4

(2014k + 1)(2018k + 1) = y^2

2014*2018k^2 + (2014+2018)k = y^2 - 1

4k(19*53*1009k + 3^2*7) = (y-1)(y+1)

Answer 5

it is easy to see that "y" must be odd

0 is not natural number

Answer 6

So how do you solve for y?

Answer 7

Does this has anything to do with modular arithmetic?

Answer 8

you can not say 0 is not a natural number as it is not universally agreed, some use 0 to be natural, some do not

Answer 9

agree

Answer 10

Which grade is the OP in? Is this grade 12 mathematics or is it university number theory?

Answer 11

he asks olympiad level problems always

Answer 12

im getting k = 945

Answer 13

if i resubtituting k = 945 to original equation. it does not works

Answer 14

looks there is no k as the solution

Answer 15

(2014k + 1)(2018k + 1) = y^2
2014*2018k^2 + (2014+2018)k + 1 = y^2
2014*2018k^2 + 4032k + 1 - y^2 = 0
D = x^2
b^2 - 4ac = x^2
(4032)^2 - 4(2014*2018)(1 - y^2) = x^2
16 + 4y^2 = x^2
x^2 - 4y^2 = 16
(x + 2y)(x - 2y) = 16
cases :
x + 2y = 16
x - 2y = 1
or
x + 2y = 8
x - 2y = 2
the 2 equations above cant be integer for y

Answer 16

it means there is no k (natural number) as the solution which make (2014k+1)(2018k+1) is a perfect square number

Answer 17

x=2016 is such value. I brute-forced it lol.

Answer 18

except x = 0 would be satisfied like you said @thomas5267 :D

Answer 19

actually i worked a solution, i might be having some algebra error
one sec..

Answer 20

i checked it by using calculator if k = 945
(2014k + 1)(2018k + 1) = y^2
(2014(945) + 1)(2018(945) + 1) = y^2
y = 1905120.0625 bla bla bla :D

Answer 21

k=2016 is one answer.

Answer 22

ah, yes k = 2016 is work.... my job is mistake also lol

Answer 23

yeah algebra mistake, k=2016 is the only solution

Answer 24

let me know if anyone is interested in the solution
it is pretty long but i found it interesting when solving

Answer 25

i think i found my mistake
i just miss one extra case :
(2014k + 1)(2018k + 1) = y^2
2014*2018k^2 + (2014+2018)k + 1 = y^2
2014*2018k^2 + 4032k + 1 - y^2 = 0
D = x^2
b^2 - 4ac = x^2
(4032)^2 - 4(2014*2018)(1 - y^2) = x^2
16 + 4y^2 = x^2
x^2 - 4y^2 = 16
(x + 2y)(x - 2y) = 16
cases :
x + 2y = 16
x - 2y = 1
or
x + 2y = 8
x - 2y = 2
or
x + 2y = 4
x - 2y = 4

the third equation is work, get solution for (x,y) = (4,0)
at least, subtitute y = 0 into :
2014*2018k^2 + (2014+2018)k + 1 = y^2
2014*2018k^2 + (2014+2018)k + 1 = 0^2
2014*2018k^2 + (2014+2018)k + 1 = 0
from here get k = 2016

Answer 26

D = x^2 guarantees rational solutions, but never integral solutions right ?

Answer 27

but if D not equal 0 be guarante the roots cant be integer :D

Answer 28

i mean D not equal x^2

Answer 29

Ahh I see, your method works !! XD

Answer 30

pretty clever !

Answer 31

i get this method from one of the member OS here, his name @mukushla
my friend from iran

Answer 32

he is my best friend on OS too

Answer 33

really, we all are friend :D

Answer 34

im not seeing him much these days
looks he got busy after starting Msc...

Answer 35

yeah, i didnt see if he was online on facebook