For consecutive odd integers a and b, a > b > 0. If a^2 + b^2 = 202, what is the value of a?

Question

I've found the answer to the question other ways, but I'm trying to understand why this method doesn't work:

a² + b² = 202

a² + b² - 202 = 0

The equation is in form a² + b² - 2ab = 0. Therefore, 2ab = 202 and ab = 101. Why is that not true and ab actually equals 99?

Answer 1

Hello Victor,

a² + b² - 2ab = 0

→ (a - b)² = 0

→ a = b.

But a > b, so they can't be the same, and so we can't solve it with that method.

I know you said you found the answer, but just wondering if your method matches mine below, or if you did something differently.

Given: a² + b² = 202

Note that b = the first consecutive odd integer, since a > b, and so

a = the second consecutive odd integer.

Then a = b + 2

→ (b + 2)² + b² = 202

→ b² + 4b + 4 + b² = 202

→ 2b² + 4b - 198 = 0

→ b² + 2b - 99 = 0

→ (b + 11)(b - 9) = 0

→ b = -11 or b = 9

Since it is given that b > 0

→ b = 9

→ a = 11

→ ab = 99.

Thank you for posting the question.

Michael Ehlers

Answer 2

x and x+2 are the integers

x^2+(x+2)^2=202

x^2+x^2+4x+4=202

2x^2+4x+4-202=0

2x^2+4x-198=0

x^2+2x-99=0

(x+11)(x-9)=0

x+11=0

x=-11, x+2=-9

x-9=0

x=9, x+2=11

a=11, b=9

disregard -9 and -11 because you want positive numbers