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Show that there are no integer solutions to m³ −n³ =3 or m³ −n³ =4.

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Final answer:

Using properties of cubes and algebraic manipulation, it's shown that there are no integer solutions for the equations m³ - n³ = 3 and m³ - n³ = 4 due to the fact that the differences in cubes of consecutive integers are larger than 3 and 4.

Step-by-step explanation:

To show that there are no integer solutions to the equations m³ − n³ = 3 and m³ − n³ = 4, we can use the properties of cubes of integers. For any two integers m and n, the difference m³ - n³ is also an integer. When considering two consecutive integers, their cubes have a difference that is larger than 3 or 4.

Let's consider the first equation m³ - n³ = 3. If n = m - 1, expanding the cubes gives us m³ - (m-1)³, which simplifies to 3m² - 3m + 1. This expression can never be equal to 3 for any integer value of m.

Now, let's consider the second equation m³ - n³ = 4. Similarly, substituting n = m - 1 and expanding the cubes results in 3m² - 3m + 1, which can't equal 4 either. Additionally, for integer solutions, there's always an integer k such that n = m - k. The difference between their cubes will grow significantly larger than 4 as m increases. Thus, no integers m and n, where m > n, can satisfy m³ - n³ = 4.

Therefore, we conclude that there are no integer solutions for both equations.

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