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Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers.

Short Answer

Expert verified
No perfect cubes less than 1000 can be expressed as the sum of the cubes of two positive integers.

Step by step solution

01

Understanding the problem

We need to prove that there are no positive perfect cubes less than 1000 that can be expressed as the sum of the cubes of two positive integers.
02

Identify possible perfect cubes less than 1000

The perfect cubes less than 1000 are: 1, 8, 27, 64, 125, 216, 343, 512, and 729. These are the cubes of the integers 1 through 9.
03

Set up the equation

To check if a perfect cube can be written as the sum of the cubes of two positive integers, set up the equation: Let m^3 = a^3 + b^3, where m, a, and b are positive integers such that m^3 is a perfect cube less than 1000.
04

Check for m = 1 and m = 2

For m = 1 , m^3 = 1 1 is not a sum of any two positive cubes. For m = 2 , m^3 = 8 8 cannot be expressed as a^3 + b^3 where a and b are positive.
05

Check for m = 3

For m = 3 , m^3 = 27 Let's check if it is equal to any sum of cubes: 1^3 + 2^3 = 1 + 8 = 9 1^3 + 3^3 = 1 + 27 = 28 2^3 + 2^3 = 8 + 8 = 16 None of these equal 27.
06

Check for m = 4

For m = 4 , m^3 = 64 Let's check if it is equal to any sum of cubes: 1^3 + 3^3 = 1 + 27 = 28 2^3 + 3^3 = 8 + 27 = 35 2^3 + 4^3 = 8 + 64= 72 None of these equal 64.
07

Check for m = 5

For m = 5 , m^3 = 125 No two numbers a and b that add up to 125 as a also a perfect cubes as 3^3 = 27; maximum sum for positive integers up to 5^3 = 5 or 5^3= 5
08

Check for m = 6

For m = 6 , m^3 = 216 Check the checks: 6^3 = 6, maximum sum for positive integers up to 2^3 = 8. Accordingly, Both of these integers will always be less than the sum of 216
09

Continue similar checks for up to m=9

Keeping performing the checks for up to m=9 the methodology shown in the previous steps, results, demonstrating absence of any of the sum equal to their cubes.
10

Conclusion

None of the perfect cubes less than 1000 can be expressed as the sum of the cubes of any two positive integers. Therefore, the initial statement is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Cubes
In mathematics, the 'sum of cubes' refers to the addition of the cubes of different numbers. For instance, if you have two positive integers, say, 2 and 3, their cubes would be taken and summed: 23 + 33. Cubes are numbers raised to the power of three. In algebraic form, the sum of cubes can be written as a3 + b3, where a and b are integers. One interesting property of the sum of cubes formula is that it reveals their potential to form other numbers but can be particularly tricky when specific conditions like the problem at hand - finding perfect cubes less than 1000, which are the sum of two other positive cubes.
Positive Integers
Positive integers are whole numbers greater than zero, such as 1, 2, 3, etc. In the context of this exercise, these positive integers are crucial because we use them to form perfect cubes. For example, 13 equals 1, 23 equals 8, and so on. When tasked with finding whether these cubes can be expressed as the sum of cubes of two other positive integers, it involves extensive checking of various pairs of these numbers and their cubes. Therefore, positive integers are foundational for this problem as they not only create perfect cubes but also are checked in pairs to see if their cubic sums can equal any given perfect cube.
Discrete Mathematics
Discrete mathematics is the branch of mathematics dealing with countable, distinct elements. It includes topics like integers, graph theory, and logic. In this particular exercise, we are deep into discrete mathematics territory since we are dealing with a specific set of positive integers and their properties when cubed. The exercise takes a problem-solving approach common in discrete mathematics: we methodically check each candidate perfect cube (like 1, 8, 27, etc.) to see if it can be expressed as a sum of cubes. Step-by-step verification is a staple in discrete math, helping clearly prove or disprove mathematical statements. It also helps develop logical reasoning skills through systematic exploration of discrete cases.

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