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# Midnight Maths. watch

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.
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
3. A perfect cube is what again?
4. (Original post by ZJuwelH)
A perfect cube is what again?
a cube of an interger
5. (Original post by Katie Heskins)
a cube of an interger
Yeah sorry I'm getting confused with a perfect number.
6. A factor is what again?

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

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

Edit: Ooopsy
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.
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.
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.
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.
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

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