Loser66
  • Loser66
For any integers a, b , prove \(35| ab(a^{12}-b^{12}\) Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Empty
  • Empty
Not sure, I just woke up, any ideas?
Loser66
  • Loser66
sorry, 35, not 36.
Loser66
  • Loser66
35 = 5*7 , hence 5| ab(a^12-b^12 ) and 7| ab(a^12 -b^12)

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Loser66
  • Loser66
since 5 is prime, hence either 5 | ab or 5| (a^12- b^12) same as 7
Empty
  • Empty
oh good point
Loser66
  • Loser66
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)
Loser66
  • Loser66
then ..... dumb! hehehe..
Loser66
  • Loser66
case 1) hence \(ab \equiv 0(mod5)\\ab\equiv 0 (mod 7)\)
Empty
  • Empty
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 \[a^{12}-b^{12} = (a^6+b^6)(a^6-b^6) = (a^6+b^6)(a^3+b^3)(a^3-b^3)\]
Empty
  • Empty
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.
mathmate
  • mathmate
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.
Loser66
  • Loser66
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.
Empty
  • Empty
oh cool @mathmate

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