The difference of two positive integers was multiplied by their sum. If possible, find two numbers such that the result is equal to 212.
(a - b)(a + b) = a^2 - b^2 = 212
find two perfect squares whose difference is 212
google "table of squares"
Reiny did this earlier
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To find the two positive integers, let's start by setting up equations based on the given problem.
Let's call the two positive integers x and y.
According to the problem, "The difference of two positive integers was multiplied by their sum." This can be expressed as:
(x - y)(x + y) = 212
Now, let's simplify this equation:
x^2 - y^2 = 212
By rearranging the equation, we get:
x^2 = y^2 + 212
Since we are looking for positive integers, we know that x must be greater than y.
Now, we can start by trying different values for y and solving for x. Here are some values of y to try:
y = 1: x^2 = 1 + 212. Since 1 + 212 = 213, there is no integer solution for x.
y = 2: x^2 = 4 + 212. Since 4 + 212 = 216, there is no integer solution for x.
y = 3: x^2 = 9 + 212. Since 9 + 212 = 221, there is no integer solution for x.
y = 4: x^2 = 16 + 212. Since 16 + 212 = 228, there is no integer solution for x.
... and so on.
By trying different values of y, we can see that there is no pair of positive integers that satisfy the given condition. Therefore, it is not possible to find two numbers whose difference multiplied by their sum is equal to 212.