Can Fermat's Little Theorem Help Disprove This Statement?

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Homework Help Overview

The discussion revolves around disproving the statement that there exist integers a, b, c, none divisible by 7, such that 7 divides the sum a^3 + b^3 + c^3. The subject area involves modular arithmetic and properties of integers under specific divisibility conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of modular arithmetic, particularly focusing on the congruences of cubes modulo 7. They discuss testing various cases for a, b, and c, and question the efficiency of their methods. Some participants also raise concerns about potential typos in modular expressions.

Discussion Status

The discussion is active, with participants sharing their reasoning and approaches. Some have suggested that there may be a more elegant solution involving Fermat's Little Theorem, while others are analyzing the patterns in the cubic residues modulo 7. There is a recognition of the need to prove that the sum cannot be divisible by 7 based on the explored cases.

Contextual Notes

Participants are working under the constraint that a, b, and c must not be divisible by 7, and they are examining the implications of this restriction on the possible values of a^3 + b^3 + c^3.

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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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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)"?
 
Samy_A said:
Is the "0(mod 3)" that appears twice a typo, and do you mean "0(mod 7)"?
sorry 0(mod7)
 
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)?
 
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?
 
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.
 
By the way: if you know Fermat's little theorem, then there is a more elegant solution.
 
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.
 
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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