Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number

Introduction

In this problem, we study the possible forms of the square of a positive integer. We will show that the square of any positive integer can never be written in the form 3m + 2, where m is a natural number.

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Question

Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Solution

Let n be any positive integer. Every positive integer can be written in one of the following three forms:

3q, 3q + 1, or 3q + 2, where q is a natural number.

We now consider the square of n in each case.

If n = 3q, then n² = (3q)² = 9q², which is divisible by 3. Hence, n² is of the form 3m.

If n = 3q + 1, then n² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which is of the form 3m + 1.

If n = 3q + 2, then n² = (3q + 2)² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, which is also of the form 3m + 1.

From the above cases, we see that the square of a positive integer can be of the form 3m or 3m + 1 only.

Hence, the square of any positive integer cannot be of the form 3m + 2.

Conclusion

Therefore, the square of any positive integer is never of the form 3m + 2, where m is a natural number.

Hence proved.

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