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Simple Number Theory

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the least positive integer N such that every integer [tex]n \geq N[/tex] can be written in the form 4a + 7b, where a,b are non-negative integers. Prove your N has this property

    2. Relevant equations

    3. The attempt at a solution
    Well, I kind of went about doing trial and error. I know 17 is not such a number, but 18,19,20,21,22 are. I figured I'd use (strong) induction. Suppose it's true for all [tex]22 \leq k \leq n[/tex]. Consider now [tex] 18 \leq n+1 -4 < n [/tex]. By the induction hypothesis n+1 -4 = 4a + 7b, so n+1 = 4(a+1) +7b, for some non-negative a,b.

    Is this correct? What if I wanted to do it more generally. That is, if gcd(x,y)=1 find N such that, for all [tex]n \geq N[/tex], n can be written as xa+yb, for non-negative a,b?
     
    Last edited: May 31, 2010
  2. jcsd
  3. May 31, 2010 #2
    That is correct.

    The general case is a little harder as we can't just use trial and error and confirm our observations. I will give you a sketch here of how to show that N=(x-1)(y-1) (especially the second part is pretty vague, but you can just ask if you're having trouble figuring it out).

    First to show that N must be at least (x-1)(y-1) assume that there exists non-negative integers a,b such that:
    ax + by = (x-1)(y-1)-1
    Then (a+1 - y)x + (b + 1)y = 0 so x | b+1, and therefore [itex]b \geq x-1[/itex] and similarly [itex]a \geq y-1[/itex]. Use these inequalities and ax+by = (x-1)(y-1) - 1 to obtain a contradiction and thereby showing N must be at least (x-1)(y-1).


    Assume without loss of generality x > y. Let r be an integer in {0,1,...,y-2}. Then
    [tex](y-2)x \leq (x-1)(y-1)+r[/tex]
    You can show that there exists an integer a in {0,1,...,y-2} such that
    [tex](y-1)(x-1) +r \equiv ax \pmod y[/tex]
    [HINT: 0x, 1x, 2x, ..., (y-2)x represents all residue classes modulo y except the residue class of -x]
    and use this to find a non-negative integer b such that (x-1)(y-1)+r = ax + by. This shows (x-1)(y-1)+r can be expressed as a linear combination of x and y with positive coefficients for r in {0,1,...,y-2}. We now only need to show it for r=y-1, but then (x-1)(y-1) + r = x(y-1).
     
  4. May 31, 2010 #3
    That last part I don't understand.
    How do you deduce from the inequality that there is an integer a with that property? I'm sorry if it should be obvious..:uhh:
     
  5. May 31, 2010 #4
    The inequality isn't actually used to find a suitable a, but rather to ensure that when we do the b we choose will be non-negative.

    0x, 1x, 2x, ..., (y-2)x, (y-1)x represent all residue classes modulo y. Therefore (x-1)(y-1)+r must be congruent to one of these modulo y. So we get that for some a in {0,1,...,y-1} we have:
    [tex](x-1)(y-1)+r \equiv ax \pmod y[/tex]
    a could not be y-1 because then we would have (using that y-1 is congruent to -1 modulo y):
    [tex]-x \equiv (y-1)x = ax \equiv (x-1)(y-1)+r \equiv -x+1+r \pmod y[/tex]
    Cancelling we get [itex]1+r \equiv 0 \pmod y[/itex], but this is a contradiction since r+1 is in {1,2,...,y-1}.
     
  6. Jun 1, 2010 #5
    Ok, thank you!
     
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