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
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To prove that the number \(10^{3n+1}\) cannot be represented as the sum of two cubes of positive integers, we will use a proof by contradiction.
Assume that \(10^{3n+1}\) can be represented as the sum of two cubes of positive integers, \(a^3\) and \(b^3\) where \(a > b\).
So, we have:
\[10^{3n+1} = a^3 + b^3 \quad (1)\]
Now, let's consider the cubes of positive integers. The cube of any positive integer can only be congruent to one of three possible residues: 0, 1, or -1 modulo 9. We can express this as:
\[a^3 \equiv 0, 1, \text{ or } -1 \pmod 9\]
\[b^3 \equiv 0, 1, \text{ or } -1 \pmod 9\]
Since \(10^{3n+1}\) is a power of 10, it can be represented as \(10\times(10^3)^n\). Let's analyze the remainders of \(10^{3n+1}\) when divided by 9:
\begin{align*}
10^{3n+1} &\equiv 10\times (10^3)^n \pmod 9 \\
&\equiv 10\times (1)^n \pmod 9 \\
&\equiv 10 \pmod 9
\end{align*}
By expressing \(10\) as \(-1 + 9\), we have:
\[10 \equiv -1 \pmod 9\]
So, \(10^{3n+1} \equiv -1 \pmod 9\).
Now, let's go back to equation (1):
\[10^{3n+1} = a^3 + b^3\]
Since the cube function modulo 9 can only produce residues of 0, 1, and -1, the sum of two cubes can only yield residues of -2, -1, 0, 1, or 2 modulo 9.
However, we have already proved that \(10^{3n+1} \equiv -1 \pmod 9\), which is none of the possible residues resulting from the sum of two cubes.
Therefore, our assumption that \(10^{3n+1}\) can be represented as the sum of two cubes of positive integers is false.
Hence, we have shown that a number \(10^{3n+1}\), where \(n\) is a positive integer, cannot be represented as the sum of two cubes of positive integers.
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 a proof by contradiction.
Assume that there exist two positive integers, a and b, such that (a^3 + b^3) = 10^(3n+1).
We know that the cube of any positive integer is congruent to either 0, 1, or -1 (mod 9). This means that for any positive integer x, x^3 is congruent to either 0, 1, or -1 when divided by 9.
Now, let's consider the possible remainders when dividing 10^(3n+1) by 9:
10 is congruent to 1 (mod 9), so 10^(3n+1) is congruent to 1^(3n+1) = 1 (mod 9).
On the other hand, when we consider a^3 + b^3, each term (a^3 and b^3) will be either 0, 1, or -1 (mod 9). The sum of two numbers congruent to 0, 1, or -1 (mod 9) can only give us a result that is also congruent to 0, 1, or -1 (mod 9). Since 10^(3n+1) is congruent to 1 (mod 9), it cannot be represented as the sum of two cubes of positive integers.
Hence, the assumption that a number of the form 10^(3n+1) can be represented as the sum of two cubes of positive integers is incorrect, establishing the proof by contradiction.