If n is a positive integer and n > 1,prove that (n) + (n-1) is a perfect sq
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To prove that (n) + (n-1) is a perfect square for n > 1, we need to show that it can be written as the square of an integer.
We can simplify the expression (n) + (n-1) as follows:
(n) + (n-1) = n + n - 1 // Removing the parentheses
Now, combining like terms, we get:
2n - 1
To show that this expression is a perfect square, we can express it as the square of an integer. Let's assume that 2n - 1 is indeed a perfect square and can be expressed as (k)^2, where k is an integer.
Therefore, we have:
2n - 1 = (k)^2 // Assume 2n - 1 is a perfect square
Adding 1 to both sides of the equation, we get:
2n = (k)^2 + 1
Now, we can observe that the right side of the equation is an odd number since (k)^2 is an even number plus 1 yields an odd number. If 2n is even, then n must also be even. However, since n > 1, n cannot be even as it would violate the given condition. Therefore, 2n must be odd, which means n must also be odd.
Now, let's rewrite k as 2m + 1, where m is an integer:
2n = (2m + 1)^2 + 1 // Substitute k = 2m + 1
Expanding the right side of the equation, we have:
2n = 4m^2 + 4m + 1 + 1
= 4m^2 + 4m + 2
Dividing both sides of the equation by 2, we get:
n = 2m^2 + 2m + 1
We can rewrite the expression on the right side as a perfect square:
n = (m + 1)^2
Therefore, we have shown that for any positive integer n > 1, (n) + (n-1) is a perfect square given that n is odd.
Note: The assumption made during the proof is known as a proof by contradiction. We started by assuming that 2n - 1 is a perfect square and showed that it leads to a contradiction. Hence, our assumption is false and (n) + (n-1) is not a perfect square when n is even.