can you answer this question in a different and more logical way than this method below:
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=2n+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 + 2kY) 22n(X2 - 2kXY + 22kY²)
=23n(X + 2kY) (X² - 2kXY + 22kY²)
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
The question is 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.
Thanks> please explain in full steps and in the most logical order
Thanks once again
The proof does look clumsy. It is however a little easier to follow if the exponents are displayed correctly. See the original version here:
http://www.jiskha.com/display.cgi?id=1247714839
Dude, it's going to be due in like three days. Do it yourself.
it's due in two days, lol
SY
To 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, we can break it down into logical steps:
Step 1: Start by assuming that you can represent 10^(3n+1) as the sum of two cubes of positive integers.
Step 2: Express the two positive integers as A and B, respectively.
Step 3: Write the sum of the cubes of A and B as A^3 + B^3.
Step 4: Notice that the sum of the cubes of two numbers can be factored using the identity (A+B)(A^2 - AB + B^2).
Step 5: Apply this identity to the sum A^3 + B^3, so it becomes (A + B)(A^2 - AB + B^2).
Step 6: Simplify the expression: (A + B)(A^2 - AB + B^2) = A^3 + B^3.
Step 7: From step 6, we have (A + B)(A^2 - AB + B^2) = A^3 + B^3 = 10^(3n+1).
Step 8: Now, analyze the possible values of A and B.
Step 9: Without losing generality, assume A = 2nX and B = 2n + kY, where X is not divisible by 2, n is a positive integer, and k is a non-negative integer.
Step 10: Substituting the values of A and B into the expression from step 7, we get:
10^(3n+1) = (A + B)(A^2 - AB + B^2)
= 2n(X + 2kY)[2^2n(X^2 - 2kXY + 2^2kY^2)]
= 2^(3n+1)(X + 2kY)(X^2 - 2kXY + 2^2kY^2)
= 2^(3n+1) * [X + 2kY)(X^2 - 2kXY + 2^2kY^2)]
Step 11: From step 10, we observe that 10^(3n+1) has a factor of 2^(3n+1) but not 2^(3n+1)+1, since X is not divisible by 2.
Step 12: Now consider the number 10^(3n+1) + 1.
Step 13: From step 12, we see that to represent 10^(3n+1) + 1 as the sum of two cubes, it would require a factor of 2^(3n+1) + 1.
Step 14: However, from step 11, we know that 10^(3n+1) does not have a factor of 2^(3n+1) + 1.
Step 15: Therefore, we can conclude that it is not possible to represent 10^(3n+1) as the sum of two cubes of positive integers.
By following these logical steps, we have proven that the number 10^(3n+1) cannot be represented as the sum of two cubes of positive integers.