Loser66
  • Loser66
Show that if a, b and c are integers with c | ab, then c| (a,c)(b, c) Please, help
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
It is easy to prove by usual way: (a, c ) = xa + yc for some x, y in Z (b, c) = sb + tc for some s, t in Z (a,c) (b,c) = (xa + yc)(sb + tc) = xasb + xatc + ycsb + yctc = ab(xs) + c ( atx ) + c (ysb) + c ( ytc) c | ab for the first term , hence c divide the sum. Problem is : my Prof wants me to use prime factorization to prove it.
Loser66
  • Loser66
My attempt: \(c = \prod p_i^{m_i}\) \(ab= \prod p_i^{s_i}\) \(c|ab \implies m_i < s_i\) Then, no where to go. :(
Loser66
  • Loser66
@ganeshie8

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anonymous
  • anonymous
I've only tried the easy way...interesting how you would use it for prime factorisation
Loser66
  • Loser66
I don't get why c | ab, implies \(c_i \geq a_i + b_i\)
ganeshie8
  • ganeshie8
Oops, it is a typo, fixed here : Let : \(a = \prod p_i^{a_i}\) \(b = \prod p_i^{b_i}\) \(c = \prod p_i^{c_i}\) then, \((a,c) = \prod p_i^{\min\{a_i,c_i\}}\) \((b,c) = \prod p_i^{\min\{b_i,c_i\}}\) \(c\mid ab \implies c_i \le a_i+b_i\) so, \(\min\{a_i,c_i\}+\min\{b_i,c_i\} \le a_i+b_i \ge c_i \blacksquare \)
Loser66
  • Loser66
Got you. Thank you so much.
ganeshie8
  • ganeshie8
np

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