# Midnight Maths.

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#1
"The integers a, b, c are non-zero and the highest common factor of a and c is 1. Given that bc(a^2 - b^2) = a(c^2 - b^4), prove that a is a perfect cube.
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16 years ago
#2
(Original post by fishpaste)
"The integers a, b, c are non-zero and the highest common factor of a and c is 1. Given that bc(a^2 - b^2) = a(c^2 - b^4), prove that a is a perfect cube.
could you use prime factors
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16 years ago
#3
A perfect cube is what again?
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16 years ago
#4
(Original post by ZJuwelH)
A perfect cube is what again?
a cube of an interger
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16 years ago
#5
(Original post by Katie Heskins)
a cube of an interger
Yeah sorry I'm getting confused with a perfect number.
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16 years ago
#6
A factor is what again?

Just kidding, but the strength of my midnight maths is a little poor right now.
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16 years ago
#7
can we wait till midnight before attempting it?
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16 years ago
#8
(Original post by elpaw)
can we wait till midnight before attempting it?
you're 10 minutes too late
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16 years ago
#9
Im afraid it is incorrect.

(a,b,c) = (12,6,72) works.

Edit: Ooopsy
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16 years ago
#10
(Original post by JamesF)
Im afraid it is incorrect.

(a,b,c) = (12,6,72) works.
But the highest common factor of a and c here is 12, the question says it has to be 1.
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16 years ago
#11
(Original post by ZJuwelH)
But the highest common factor of a and c here is 12, the question says it has to be 1.
Oh yea, forgot that condition.
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16 years ago
#12
Divide through by a initially to give
c^2 -b^4 = abc - cb^3/a

Either
a|c => a=c=1
Or
a|b^3 therefore b^3 = an (for some integer n)

Divide by b initially
c(a^2 - b^2) = ac^2/b - ab^3 (ac^2/b must be an integer)

Either
b|c^2 => b^3|c^6 => a|b^3|c^6 => a|c^6, gcd(a,c) = 1, implies a=c=1 => b=1 by substituting for a and c.

Or
b|a therefore a = bm (for some integer m)

If we divide through by bc initially we have
a^2-b^2 = ac/b - ab^3/c = mc - ab^3/c (since a = bm)

but b = a/m, therefore a^2-b^2 = mc - a(a/m)^3/c = mc - a^4/cm^3

Therefore c|a^4 ==> a=c=1 and therefore b=1 since b|a

So all cases give the solution (1,1,1), and a = 1 is a perfect cube.
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16 years ago
#13
I think James has done this already but here is one method. If bc(a^2 - b^2) = a(c^2 - b^4) we have a quadratic in a:

bca^2 - (c^2-b^4)a - cb^3 = 0

Solving using the quadratic formula gives (I don't want to write all the algebra out here),

a = c/b or -b^3/c

Therefore, the equation is true only when one of the following equations are satisfied

(1) ab = c
(2) ac = -b^3

Considering (1), it is clear that since c a multiple of a, the highest common factor of a and c is a. Therefore, since the hcf of a and c is 1, the only solution of (1) is a=1, which is a perfect cube. Since a must be an integer, the only solutions for a in equation (2) are a=±1,±b,±b^2,±b^3. But, if a=±b,±b^2 it must be the case that c=±b^2,±b which means that b would be a common factor of a and c, which means that b must be 1 because the hcf of a and c is 1. So, in the cases a=±b,±b^2 we have shown that b must equal 1, so a=±1 which are perfect cubes. The remaining cases are a=±1,±b^3 which are perfect cubes.
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16 years ago
#14
(Original post by ZJuwelH)
Yeah sorry I'm getting confused with a perfect number.

i think u must be good at maths amm can u help me with introduction in number grid 0
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