The number feared by Pythagoreans, since it lies halfway between the only two integers that can be both the perimeter and the area of the same rectangle?

http://www.yellowpigs.net/yellowpigs/YP_seventeen

The number feared by Pythagoreans, since it lies halfway between the only two integers that can be both the perimeter and the area of the same rectangle?

A 3x6 rectangle has an area and perimeter of 18 and a 4x4 rectangle has an area and perimeter of 16.

Could 17 be the number you seek?

bob

To find the number feared by Pythagoreans, we need to first understand the concept of the perimeter and area of a rectangle.

The perimeter of a rectangle is calculated by adding up the lengths of all four sides of the rectangle. If we let 'l' represent the length and 'w' represent the width of the rectangle, then the perimeter (P) can be calculated as:

P = 2(l + w)

The area of a rectangle is calculated by multiplying the length and width of the rectangle. The formula for finding the area (A) is:

A = l * w

Now, according to the question, we are looking for a number that can be both the perimeter and the area of the same rectangle. In other words, we are looking for a value of 'l' and 'w' that satisfies the equations:

2(l + w) = l * w

Simplifying this equation, we get:

2l + 2w = lw

Rearranging the equation, we get:

lw - 2l - 2w = 0

To solve this equation, we need to use a quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for 'x' are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation lw - 2l - 2w = 0, we can treat 'l' as 'x' and 'w' as 'y' to rewrite the equation in the form of a quadratic equation:

xy - 2x - 2y = 0

Using the quadratic formula, we can find the values of 'l' and 'w' such that the equation is satisfied. Now, let's substitute 'l' with 'x' and 'w' with 'y' in the quadratic formula:

x = (-(-2) ± √((-2)^2 - 4(1)(-2y))) / (2(1))

Simplifying further:

x = (2 ± √(4 + 8y)) / 2

x = 1 ± √(1 + 2y)

Now, let's analyze the term inside the square root, (1 + 2y).

Since we are looking for an integer value of 'l' and 'w', (1 + 2y) must be a perfect square. This means that 1 + 2y is the square of an integer.

Now, we can list down the possible perfect squares that can be expressed as (1 + 2y):

1 + 2y = 1, 4, 9, 16, 25, 36, ...

From this list, it is clear that the only two consecutive numbers from this series are 9 and 16. These numbers satisfy the equation (1 + 2y) = 9, which gives 'y' a value of 4, and (1 + 2y) = 16, which gives 'y' a value of 7.

Now, substituting 'y' back into the expression for 'x', we get:

x = 1 ± √(1 + 2y)
x = 1 ± √(1 + 2*4) , which gives us x = 3 or x = -1 (not a valid length)

Since we are looking for positive values of 'l' and 'w', the only valid solution is x = 3. Therefore, the number that is feared by Pythagoreans, which lies halfway between the only two integers that can be both the perimeter and the area of the same rectangle, is 3.