Solution 76: Polygonal Numbers and GCD's

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  • Published on Jan 24, 2019
  • Let's use properties of divisors to gain information about b_n.
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Comments • 12

  • er3z
    er3z 7 months ago

    yes one thing i want to ask you

    what

  • Max Haibara
    Max Haibara 7 months ago +1

    wait... you said that if b|a and b|c, then b|ax+cy for ANY integer x,y. But then you only use SPECIFIC value of x,y which is n+1 and n-1 to prove it. How about the other value of x,y?

    • Easy Mathematics
      Easy Mathematics 7 months ago

      @Max Haibara The fact he uses is easy to proof. There is no problem.
      if a | b and a | c, then a | xb + yc for any x,y.
      In words:
      If a divides two integers, then a divides every linear combination.
      This fact is easy to proof.
      So he choosed a specific one. For purpose. That's all and everything is correct.
      I don't understand your problem.

    • Max Haibara
      Max Haibara 7 months ago

      @Easy Mathematics but not all ax+cy will be equal to 1, at least we need to prove that all x,y will make ax+cy=1
      If there is a value of x,y that makes ax+cy!=1, then it's not valid

    • triton62674
      triton62674 7 months ago

      The theorem he uses supposes it's already true for any integer and he then uses a specific one

    • Easy Mathematics
      Easy Mathematics 7 months ago +1

      If it is valid for any then it is valid for specific.

  • adandap
    adandap 7 months ago +5

    That is a sophisticated and elegant argument. I really have to stop just charging ahead with algebra!

  • Sudheer Thunga
    Sudheer Thunga 7 months ago +2

    So...the theorem you stated at 4:51 states that when c is a divisor of gcd(a,b)... c is also a divisor of a and b.

    • Sudheer Thunga
      Sudheer Thunga 7 months ago

      Oh... thank you

    • LetsSolveMathProblems
      LetsSolveMathProblems  7 months ago

      Yes, you are correct. As a fun fact, the converse is true, as well: If c is a divisor of a and b, it is a divisor of gcd(a,b). The converse is slightly harder to prove.

  • mark erena
    mark erena 7 months ago

    First

  • {GLD} p0pc0r0n
    {GLD} p0pc0r0n 7 months ago +10

    I wanna see let’s solve math problems vs blackpenredpen