# Midnight Maths.

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"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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#2

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"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.

**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.

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#5

(Original post by

a cube of an interger

**Katie Heskins**)a cube of an interger

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#6

A factor is what again?

Just kidding, but the strength of my midnight maths is a little poor right now.

Just kidding, but the strength of my midnight maths is a little poor right now.

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#8

(Original post by

can we wait till midnight before attempting it?

**elpaw**)can we wait till midnight before attempting it?

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#10

(Original post by

Im afraid it is incorrect.

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

**JamesF**)Im afraid it is incorrect.

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

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#11

(Original post by

But the highest common factor of a and c here is 12, the question says it has to be 1.

**ZJuwelH**)But the highest common factor of a and c here is 12, the question says it has to be 1.

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#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.

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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#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

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.

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/cTherefore, 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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#14

(Original post by

Yeah sorry I'm getting confused with a perfect number.

**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

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