Question

For any integers a, b , prove \(35| ab(a^{12}-b^{12}\)
Please, help

Answer 1

Not sure, I just woke up, any ideas?

Answer 2

sorry, 35, not 36.

Answer 3

35 = 5*7 , hence
5| ab(a^12-b^12 ) and 7| ab(a^12 -b^12)

Answer 4

since 5 is prime, hence either 5 | ab or 5| (a^12- b^12)
same as 7

Answer 5

oh good point

Answer 6

Now, match them up
case 1) 5| ab and 7| ab
case2) 5| ab and 7| (a^12 - b^12)
case 3) 5| (a^12- b^12) and 7| ab
case 4) 5| (a^12 -b^12 ) and 7| (a^12- b^12)

Answer 7

then ..... dumb! hehehe..

Answer 8

case 1) hence \(ab \equiv 0(mod5)\\ab\equiv 0 (mod 7)\)

Answer 9

Oh here's something I just noticed, this could be useful to us, idk if that last part can simplify down further, diference of cubes I think can simplify but I forget off the top of my head how

\[a^{12}-b^{12} = (a^6+b^6)(a^6-b^6) = (a^6+b^6)(a^3+b^3)(a^3-b^3)\]

Answer 10

here we go,
\[ab(a^{12}-b^{12}) = ab(a^6+b^6)(a^3+b^3)(a^2+ab+b^2)(a-b)\]

Like I think there's a elegant solution, even though we can often sorta brute force mod arithmetic problems by cases I'm usually too lazy to do that, unfortunately I don't know how to proceed past this.

Answer 11

actually it can factor further, giving
\(ab(a^{12}-b^{12})=ab(a-b)(b+a)(b^2+a^2)(b^2-a*b+a^2)(b^2+ab+a^2)(b^4-a^2b^2+a^4)\)
giving 8 factors altogether.

Answer 12

I have class in 15 minutes. I am sorry for not being here to get help. However, I would like to let it open so that if someone can help, it is perfect.

Answer 13

oh cool @mathmate