Prove that 5 does not divide (a^3) +(a^2) +1

Prove that 5 does not divide (a^3) +(a^2) +1 . This is the question , I'm really struggling with this could anybody check if this is a valid proof I have completed involving mod5 ?

"If (a^3) +(a^2) +1 was divisible by 5 then
(a^3) +(a^2) +1= 0(mod5)
(a^3) +(a^2) = 1(mod5)

The equivalence classes of mod5 are 0,1,2,3 and 4
Subsitiuting these in place of a

(0)+(0) =1(mod5) ,
((1)+(1) =1(mod5)
(8) +(4)=1(mod5)
(27) +(9) = 1(mod5)
(64) +(16)=1(mod5)
[Note all the equals signs above are meant to represent congruency]"

I'm struggling on how to word the next part . If what I've done so far is correct then since none of the numbers on the LHS are equal to 1(mod5) then 5 does not divide the equation .

Hopefully this is almost at a valid proof . Thanks for any help :smile:

Reply 1

IDGAF

Original post by Matt5422

Prove that 5 does not divide (a^3) +(a^2) +1 . This is the question , I'm really struggling with this could anybody check if this is a valid proof I have completed involving mod5 ?

"If (a^3) +(a^2) +1 was divisible by 5 then
(a^3) +(a^2) +1= 0(mod5)
(a^3) +(a^2) = 1(mod5)

The equivalence classes of mod5 are 0,1,2,3 and 4
Subsitiuting these in place of a

(0)+(0) =1(mod5) ,
((1)+(1) =1(mod5)
(8) +(4)=1(mod5)
(27) +(9) = 1(mod5)
(64) +(16)=1(mod5)
[Note all the equals signs above are meant to represent congruency]"

I'm struggling on how to word the next part . If what I've done so far is correct then since none of the numbers on the LHS are equal to 1(mod5) then 5 does not divide the equation .

Hopefully this is almost at a valid proof . Thanks for any help :smile:

The step that I've put in bold is wrong. It should be a^3 + a^2 = -1 = 4 (mod 5). Therefore none of the equivalence classes should be congruent to 4 (not 1); and none of them are.

(edited 13 years ago)

Reply 2

Raiden10

17

Original post by Matt5422

Prove that 5 does not divide (a^3) +(a^2) +1 . This is the question , I'm really struggling with this could anybody check if this is a valid proof I have completed involving mod5 ?

"If (a^3) +(a^2) +1 was divisible by 5 then
(a^3) +(a^2) +1= 0(mod5)
(a^3) +(a^2) = 1(mod5)

The equivalence classes of mod5 are 0,1,2,3 and 4
Subsitiuting these in place of a

(0)+(0) =1(mod5) ,
((1)+(1) =1(mod5)
(8) +(4)=1(mod5)
(27) +(9) = 1(mod5)
(64) +(16)=1(mod5)
[Note all the equals signs above are meant to represent congruency]"

I'm struggling on how to word the next part . If what I've done so far is correct then since none of the numbers on the LHS are equal to 1(mod5) then 5 does not divide the equation .

Hopefully this is almost at a valid proof . Thanks for any help :smile:

0+0+1=1 (mod 5)
1+1+1=3 (mod 5)
8+4+1=13=3 (mod 5)
27+9+1=37=2 (mod 5)
64+16+1=81=1 (mod 5)

Reply 3

Raiden10

17

Original post by Raiden10

0+0+1=1 (mod 5)
1+1+1=3 (mod 5)
8+4+1=13=3 (mod 5)
27+9+1=37=2 (mod 5)
64+16+1=81=1 (mod 5)

I'm not sure exactly how careful you are supposed to be, but that pretty much settles it. You are looking for 0's on the RHS, and since there aren't any, that completes the proof.