Answer to Question #87045 in Algebra for Katherinne

Question

The sum of the squares of two consecutive is 1201. What are the integers?

Answer 1

Consecutive integers are n and n+1 for some integer n. The the problem can be formulated as follows:

Find such integer n that n^2 + (n+1)^2 = 1201.

Solution:

First, expand the second bracket: n^2 + (n^2 + 1 + 2n) = 1201.

Then group the addends to form quadratic equation: 2n^2 + 2n - 1200 = 0. It can be reduced to n^2 + n - 600 = 0 by dividing by 2. Then we only need to solve this equation.

Discriminant D = 1^2 - 4*1*(-600) = 2401 = 49^2.

Then the roots will be:

n1 = (-1 - √(D))/(2*1) = -50/2 = -25,

n2 = (-1 + √(D))/(2*1) = 48/2 = 24.

If we are looking only for positive integers the answer is 24, 25. If negatives are allowed, the -25, -24 are also in the answer. From the solution above it is obvious that only such answers are possible, since a quadratic equation can have at most 2 real roots.