Are there two consecutive square numbers that differ by 3721? explain.please
Are you saying that the difference between x^2 and (x+1)^2 is 3721?
Solve Leo's equation
To find out if there are two consecutive square numbers that differ by 3721, we can use algebra to solve the problem.
Let's assume that the first square number is n^2. Since we want to find two consecutive square numbers, the second square number can be represented as (n+1)^2.
According to the problem, the difference between these two squares is given as 3721. This can be expressed as:
(n+1)^2 - n^2 = 3721
Expanding the equation:
n^2 + 2n + 1 - n^2 = 3721
Simplifying the equation further:
2n + 1 = 3721
Now, we can isolate the variable n by subtracting 1 from both sides:
2n = 3721 - 1
2n = 3720
Next, divide both sides by 2 to solve for n:
n = 3720 / 2
n = 1860
So, the value of n is 1860. Now, substituting this value into our expression for the first square number, we find that the first square number is 1860^2.
To find the second square number, we use (n+1)^2:
(1860 + 1)^2 = 1861^2
Therefore, the two consecutive square numbers that differ by 3721 are 1860^2 and 1861^2.