The product of two consecutive positive integers is added to the larger of the two integers. Prove that the result is always a square number.
Thank you for your help.
1st number = X.
2nd number = X+1.
(x+1) + x(x+1) = x+1 + x^2+x = x^2 + 2x + 1 = (x+1)(x+1) = (x+1)^2.
To prove that the result is always a square number, we need to demonstrate that for any two consecutive positive integers, their product added to the larger integer will yield a perfect square.
Let's take two consecutive positive integers: n and n+1.
The product of these two integers is: n * (n+1) = n^2 + n.
We will now express this sum as a square of another number.
Let's consider the number (n + 1/2). Squaring it gives us: (n + 1/2)^2 = n^2 + (1/2)^2 + 2 * n * (1/2).
Expanding the square, we get: n^2 + 1/4 + n = n^2 + n + 1/4.
Now, let's analyze the expression we obtained for the square of (n + 1/2) and compare it to the given expression for the product of the two consecutive integers, n^2 + n.
By comparing the two expressions, we can see that n^2 + n + 1/4 is equal to the given n^2 + n expression. Therefore, the sum of the product and the larger of the two integers is always a perfect square, specifically (n + 1/2)^2.
This demonstrates that, for any two consecutive positive integers, the result of the product of the integers added to the larger integer is always a square number.