# Elementary math proof

1. May 7, 2016

### lolo94

1. The problem statement, all variables and given/known data
. Disprove the following statement: There exists integers a, b, c, none divisible by 7, such that 7|a^3 + b^3 + c^3
2. Relevant equations

3. The attempt at a solution
if 7|a^3 + b^3 + c^3, then a^3 + b^3 + c^3 is congruent to 0(mod 7)

if a,b,c are none divisible by 7 then I just work out the cases for 1,2,3,4,5,6 and show that there is no way to get to a^3 + b^3 + c^3 is congruent to 0(mod 7).

Is that right?

Is there an easier way to do it cause mine is very inefficient.

Last edited: May 7, 2016
2. May 7, 2016

### Samy_A

Is the "0(mod 3)" that appears twice a typo, and do you mean "0(mod 7)"?

3. May 7, 2016

### lolo94

sorry 0(mod7)

4. May 7, 2016

### Samy_A

You don't give details on how you did it, so maybe what follows is moot.

Let a be an integer not divisible by 7, and r = a (mod 7).
What is the relation between a³ (mod 7) and r³ (mod 7)?
What are the possible values for r³ (mod 7)?

5. May 7, 2016

### lolo94

Yes I have the same approach a is congruent to 1,2,3,4,5,6 mod 7. Same thing for b and c. If I cube the congruence for each case, I show that there is no way you will get to a^3+b^3+c^3 congruent to 0(mod 7)

a^3 congruent to 1,8,27,64,125 mod 7

Ohhhh I got it so my last expression is equivalent to a^3 is congruent to 1 mod 7

a^3 congruent to (1 mod 7)
b^3 congruent to (1 mod7)
c^3 congruent to 1 (mod7)

a^3+b^3+c^3 is congruent to 3 mod 7
right?

6. May 7, 2016

### Samy_A

Not quite.
1 = 1 (mod 7)
8 = 1 (mod 7)
but 27 = 6 (mod 7)
and so on.
You also forgot 6³ = 216.

But yes, there is a pattern.

7. May 7, 2016

### Samy_A

By the way: if you know Fermat's little theorem, then there is a more elegant solution.

8. May 7, 2016

### lolo94

yeah I forgot that one xD, but yours state that 36 is congruent to 1 mod 7 and not 0 mod 7. If we try every case we should end up seeing that it's never congruent to 0 mod 7. Right?

Everyone talks about fermat's little theorem. They always suggest me to use that in nearly 70% of the problems that I do XD. I will look at it.

9. May 7, 2016

### Samy_A

Not sure I understand what you say here.
If you just inspect the values of 1³ (mod 7), 2³ (mod 7), ... , 6³ (mod 7), you will notice a clear pattern in the possible values.
Now, you have to add the numbers (a³ + b³ + c³), and prove that that number is not divisible by 7. Once you have noticed the pattern mentioned above, you are left with a small number of cases to inspect.

10. May 7, 2016

### lolo94

what I am saying is that

a^3 is congruent to 6,1 mod 7
b^3 is congruent to 6,1 mod 7
c^3 is congruent to 6,1 mod 7

so my point is that you can never add up these numbers to compute a multiple of 7. Examole, 6,6,6,.....6,1,1,......6,1,6

11. May 7, 2016

### Samy_A

Yes, that is correct. You basically have 4 cases: 1 1 1, 1 1 6, 1 6 6, 6 6 6, and none adds up to a multiple of 7.

12. May 7, 2016

### lolo94

thanks!!!!!!. I will study the fermat's little theorem. It's too famous.