Can Fermat's Little Theorem Help Disprove This Statement?

In summary, if 7|a^3+b^3+c^3, then a^3+b^3+c^3 is congruent to 0(mod 3). 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 3).
  • #1
lolo94
17
0

Homework Statement


. Disprove the following statement: There exists integers a, b, c, none divisible by 7, such that 7|a^3 + b^3 + c^3

Homework Equations

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.
 
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  • #2
lolo94 said:

Homework Statement


. Disprove the following statement: There exists integers a, b, c, none divisible by 7, such that 7|a^3 + b^3 + c^3

Homework Equations

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 3)

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 3).

Is that right?

Is there an easier way to do it cause mine is very inefficient.
Is the "0(mod 3)" that appears twice a typo, and do you mean "0(mod 7)"?
 
  • #3
Samy_A said:
Is the "0(mod 3)" that appears twice a typo, and do you mean "0(mod 7)"?
sorry 0(mod7)
 
  • #4
Your method is not wrong.
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
Samy_A said:
Your method is not wrong.
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)?
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 7Ohhhh 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
lolo94 said:
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 7Ohhhh 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?
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
By the way: if you know Fermat's little theorem, then there is a more elegant solution.
 
  • #8
Samy_A said:
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.
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
lolo94 said:
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?
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
Samy_A said:
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.
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
lolo94 said:
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
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
Samy_A said:
Yes, that is correct.
thanks!. I will study the fermat's little theorem. It's too famous.
 

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