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.
let the 3 terms be defined in the usual way of a a GS
a, ar, and ar^2
so
we have to show that
(a^2 + a^2 r^2 + a^2 r^4) / (a + ar + ar^2) divides evenly
that is, our denominator should cancel
(a^2 + a^2 r^2 + a^2 r^4) / (a + ar + ar^2)
= a^2(1 + r^2 + r^4)/( a(1+r + r^2) )
= a(1 + r^2 + r^4)/(1+r + r^2)
now 1+r^2 + r^4 is a GS , with a=1 , and common ratio of r^2
Sum(3) = 1( (r^2)^3 - 1)/(r^2 - 1)
= (r^6 - 1)/(r^2 - 1)
and 1 + r + r^2 is a GS, with a=1 and common ratio of r
sum(3) = (r^3-1)/(r-1)
so ....
a(1 + r^2 + r^4)/(1+r + r^2)
=a(r^6 - 1)/(r^2-1) รท ( (r^3 - 1/(r-1) )
= a(r^6 - 1)/(r^2-1) * (r-1)/(r^3-1)
= a(r^3-1)(r^3+1)(r-1)/( (r+1)(r-1)(r^3-1) )
= a(r^3 + 1)/(r+1)
= a(r+1)(r^2 - r + 1)/(r+1) ---- using sum of cubes factoring
= a(r^2 - r + 1) , no divisors left at the bottom
Hope somebody can come up with a shorter way.
CHECK:
eg. suppose we have 7, 28, 112
here a=7 , r = 4
a^2 + b^2 + c^2 = 7^2 + 28^2 + 112^2 = 13377
a+b+c = 7+28+112 = 147
13377/147 = 91
according to my formula the quotient should be
a(r^2 - r+1)
= 7(16-4+1) = 91 YEAHHHH!
Looks good to me. That was going to be my approach until I saw it so ably demonstrated...
To prove that the number a^2 + b^2 + c^2 is divisible by the number a + b + c, we can use algebraic manipulation and exploit the properties of geometric progression.
Let's start by assuming that a, b, and c are three integers in a geometric progression. This means that:
b = ar
c = ar^2
where 'r' is the common ratio of the geometric progression.
Now, we can rewrite the expression a^2 + b^2 + c^2 as follows:
a^2 + b^2 + c^2 = a^2 + (ar)^2 + (ar^2)^2
= a^2 + a^2r^2 + a^2r^4
Factoring out 'a^2' from each term, we have:
a^2 + a^2r^2 + a^2r^4 = a^2(1 + r^2 + r^4)
Now, let's consider the expression a + b + c:
a + b + c = a + ar + ar^2
= a(1 + r + r^2)
Comparing the expressions for a^2 + b^2 + c^2 and a + b + c, we notice that a^2 + b^2 + c^2 can be expressed as a^2 multiplied by (1 + r^2 + r^4), while a + b + c can be expressed as a multiplied by (1 + r + r^2).
To prove that a^2 + b^2 + c^2 is divisible by a + b + c, it is sufficient to show that (1 + r^2 + r^4) is divisible by (1 + r + r^2).
To do this, we can multiply both sides of the expression (1 + r + r^2) by (1 - r), which is the common ratio of the geometric progression:
(1 + r^2 + r^4)(1 - r) = 1 - r^3
Using the property of geometric progression again, we know that r^3 = 1, since a, b, and c are in a geometric progression. Therefore:
(1 + r^2 + r^4)(1 - r) = 1 - r^3
= 1 - 1
= 0
Since the expression (1 + r^2 + r^4)(1 - r) evaluates to zero, we can conclude that (1 + r^2 + r^4) is divisible by (1 + r + r^2). Therefore, a^2 + b^2 + c^2 is divisible by a + b + c.
Hence, we have proven that the number a^2 + b^2 + c^2 is exactly divisible by the number a + b + c when a, b, and c are three integers in geometric progression.