Prove that a number 10^(3n+1) , where n is a positive integer, cannot be represented as the sum of two cubes of positive integers.

Question

Prove that a number 10^(3n+1) , where n is a positive integer, cannot be represented as the sum of two cubes of positive integers.

thanx

Answer 1

We will examine the sum of cubes of two numbers, A and B. Without losing generality, we will further assume that

A=2ⁿX and
B=2^n+kY
where
X is not divisible by 2
n is a positive integer and
k is a non-negative integer.

A³+B³
=(A+B)(A²-AB+B²)
=2ⁿ(X + 2^kY) 2²ⁿ(X² - 2^kXY + 2^2kY²)
=2³ⁿ(X + 2^kY) (X² - 2^kXY + 2^2kY²)
Thus A³+B³ has a factor 2³ⁿ, but not 2³ⁿ⁺¹ since X is not divisible by 2.
Since 10³ⁿ⁺¹ requires a factor of 2³ⁿ⁺¹, we conclude that it is not possible that
10³ⁿ⁺¹=A³+B³

Answer 2

Nice Answer, But Please Try To Use (Mod)

That Way Is Easier

Answer 3

Hey, Your ANSWER is corrupt, cause it doesnt really explain anything! Try to make it more clear.

SY

Answer 4

To prove that a number of the form 10^(3n+1) cannot be represented as the sum of two cubes of positive integers, we can use the method of contradiction.

Let's assume that there exist positive integers a and b such that a^3 + b^3 = 10^(3n+1).

Now, we know that the cubic residues modulo 10 are {0, 1, 8, 7, 4, 3, 6, 9, 2, 5}. This means that any perfect cube will have a remainder of one of these values when divided by 10.

Considering this, let's analyze the possible remainders of a^3 and b^3 when divided by 10:

1. If a^3 ≡ 0 (mod 10), then a ≡ 0 (mod 10). In this case, a^3 will be divisible by 1000 (10^(3n+1)). However, b^3 is a perfect cube and cannot be zero modulo 10 since its remainder must fall within the set of cubic residues mentioned above.

2. If a^3 ≡ 1 (mod 10), then a ≡ 1 (mod 10). In this case, a^3 ≡ 1 (mod 10). However, 10^(3n+1) ≡ 0 (mod 10) since it ends in zero. Therefore, a^3 + b^3 cannot be equivalent to 10^(3n+1).

3. If a^3 ≡ 8 (mod 10), then a ≡ 2 (mod 10). In this case, a^3 ≡ 8 (mod 10). However, 10^(3n+1) ≡ 0 (mod 10) since it ends in zero. Therefore, a^3 + b^3 cannot be equivalent to 10^(3n+1).

4. If a^3 ≡ 7 (mod 10), then a ≡ 3 (mod 10). In this case, a^3 ≡ 7 (mod 10). However, 10^(3n+1) ≡ 0 (mod 10) since it ends in zero. Therefore, a^3 + b^3 cannot be equivalent to 10^(3n+1).

We can continue this analysis for each remainder in the cubic residues modulo 10. In each case, we will find that a^3 + b^3 is not equivalent to 10^(3n+1), which contradicts our initial assumption.

Hence, we have proven that a number of the form 10^(3n+1) cannot be represented as the sum of two cubes of positive integers.