Question

If all numbers are defined to be positive integers of the form 4n + 1 (i.e. 1,5,9,13,17, etc), show that the product of two of these numbers is also of the form 4n + 1.

Answer 1

=(4n+1)*(4n+1)
=16n^2+4n+4n+1
=16n^2+8n+1
=4(4n^2+2n)+1
hmmm I wonder if this is correct

Answer 2

Hence its in the form 4k+1 where k=4n^2+2n

Answer 3

That's where I was trying to go with this too, but it didn't seem to go anywhere.

Answer 4

Seems like that's what should be done though; I just can't figure out how to bring it back around to 4n+1 without trivially dividing it by 4n+1.

Answer 5

Why u bringing it back to its original state???

Answer 6

Because the point is to show that the product is of the same form...?

Answer 7

(4n+1)*4(n+1)=(4n+1)^2
=16n^2+4n+4n+1
=16n^2+8n+1
=4(4n^2+2n)+1
=4k+1 where k=4n^2+2n
Hence we see that the product of of 4n+1 still retains the form

Answer 8

Ah, yes! It was staring me in the face the entire time! Thanks!

Answer 9

I can tag someone else for their opinion if ud like :)

Answer 10

@dan815

Answer 11

this is exactly how we prove these... looks good to me !

Answer 12

Thanks :)
I doubt myself

Answer 13

@ikram002p

Answer 14

oh nooo ikram is here <3

Answer 15

its correct Rachel :P

Answer 16

Thanks :D

Answer 17

are u working on mod ganesh ?
or some other idea ?

Answer 18

btw rachel ur profile pic freaks me out xD

Answer 19

lmao Ik same here .... I gotta change it

Answer 20

no, the original proof is good enough :)

Answer 21

wait a second, i see a blunder

Answer 22

sh!t :(

Answer 23

you're assuming both numbers are same in ur proof, check once

Answer 24

its ok she can change and make it m
anyway to see this in other way.(lazy mod)

let 4m+1=1 mod 4
4n+1= 1 mod 4
(4n+1)(4m+1)= 1*1 mod 4
(4n+1)(4m+1)= 1 mod 4
(4n+1)(4m+1)= 4 k +1

Answer 25

Neat :)

Answer 26

(4n+1)*(4m+1)
=4n(4m+1) + 1(4m+1)
=4(4mn+n+m) + 1
=4K + 1

Answer 27

hehe

Answer 28

Yup I forgot abt that :)
Thanks guys