If a, b, and c are three integers in geometric progression, prove that the number a^2+b^2+c^2 is exactly divisible by the number a+b+c.
To prove that the number a^2 + b^2 + c^2 is exactly divisible by the number a + b + c, we can use the concept of geometric progression and algebraic manipulation.
Let's assume that the common ratio of the geometric progression is 'r'. Therefore, we can express a, b, and c as follows:
a = x
b = x * r
c = x * r^2
Now, let's calculate the sum of the three integers, a + b + c:
a + b + c = x + x * r + x * r^2 = x(1 + r + r^2)
Next, let's calculate the square of each of the three integers:
a^2 = x^2
b^2 = (x * r)^2 = x^2 * r^2
c^2 = (x * r^2)^2 = x^2 * r^4
Now, let's calculate the sum of the squares, a^2 + b^2 + c^2:
a^2 + b^2 + c^2 = x^2 + x^2 * r^2 + x^2 * r^4 = x^2(1 + r^2 + r^4)
We need to prove that a^2 + b^2 + c^2 is divisible by a + b + c:
a^2 + b^2 + c^2 = x^2(1 + r^2 + r^4)
a + b + c = x(1 + r + r^2)
To prove the divisibility, we need to show that there exists an integer 'k' such that: a^2 + b^2 + c^2 = k * (a + b + c).
Let's divide a^2 + b^2 + c^2 by a + b + c:
(a^2 + b^2 + c^2) / (a + b + c) = (x^2(1 + r^2 + r^4)) / (x(1 + r + r^2))
= x(1 + r^2 + r^4) / (1 + r + r^2)
We can simplify this expression by factoring x out of the numerator:
(a^2 + b^2 + c^2) / (a + b + c) = x(1 + r^2 + r^4) / (1 + r + r^2)
= x(r^0 + r^2 + r^4) / (1 + r + r^2)
Notice that r^0 = 1, so we can rewrite the expression as:
(a^2 + b^2 + c^2) / (a + b + c) = x(1 + r^2 + r^4) / (1 + r + r^2)
= x(r^0 + r^2 + r^4) / (1 + r + r^2)
= x(r^0 + r^2 + r^2 * r^2) / (1 + r + r^2)
Now, we can factor r^2 out of the denominator:
(a^2 + b^2 + c^2) / (a + b + c) = x(r^0 + r^2 + r^2 * r^2) / (1 + r + r^2)
= x(r^0 + r^2(1 + r^2)) / (1 + r + r^2)
Since r is a common ratio, we know that r ≠ -1. Therefore, we can cancel out the (1 + r + r^2) in both the numerator and the denominator:
(a^2 + b^2 + c^2) / (a + b + c) = x(r^0 + r^2(1 + r^2)) / (1 + r + r^2)
= x(r^0 + r^2(1 + r^2)) / (1 + r + r^2)
= x(r^0 + r^2(1 + r^2)) / (1 + r + r^2)
Now, we can see that the numerator x(r^0 + r^2(1 + r^2)) is divisible by x, which means that the entire expression is divisible by (a + b + c). Thus, we have shown that a^2 + b^2 + c^2 is exactly divisible by a + b + c.
Therefore, the proof is complete.