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Math Help - System of two equations with three integer unknowns

  1. #1
    Super Member Bacterius's Avatar
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    System of two equations with three integer unknowns

    Hi all,
    here is my problem. I have the following system :

    S \left\{\begin{array}{cc}a^2 - ab + n = 0<br />
\\a^2 - ac - n = 0\end{array}\right.

    Where a, b and c are integer unknowns and n is any integer value (known)

    Is there any general way to solve such equations quickly (for all three unknowns) ? Thanks a lot !
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  2. #2
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    Quote Originally Posted by Bacterius View Post
    Hi all,
    here is my problem. I have the following system :

    S \left\{\begin{array}{cc}a^2 - ab + n = 0<br />
\\a^2 - ac - n = 0\end{array}\right.

    Where a, b and c are integer unknowns and n is any integer value (known)

    Is there any general way to solve such equations quickly (for all three unknowns) ? Thanks a lot !
    The first thing I notice is that adding the two equations gives 2a^2- a(b+c)= 0. One obvious solution to that is a= 0. But that gives a^2- ab+ n= n= 0 and [tex]a^2- ac- n= n= 0[/itex]. If n= 0, then a= 0 , b, c any integers satisfy the equations.

    If a is not 0, we can divide through by it and get a- (b+c)= 0 so a= b+c. Putting that into the first equation we get (b+c)^2- (b+c)b+ n= b^2+ 2bc+ c^2- b^2- bc+ n = bc+ c^2+ n= 0 so c^2= bc- n.

    Putting a= b+ c into the last eqaution gives (b+c)^2- (b+c)c- n= 0 or b^2+ 2bc+ c^2- bc- c^2- n= b^2+ bc- n= 0 and, now, c^2= bc-n. That means we must have b^2= c^2 and so c= \pm b.

    If c= -b, we have, from a-(b+c)= a= 0 again.

    If c= b, a-(b+c)= 0 becomes a- 2b= 0 so a= 2b. Given any integer, k, take a= 2k, b= c= k. The equation a^2- ab+ n= 0 becomes 4k^2- k^2+ n= 3k^2+ n= 0 .

    If n> 0, that is impossible. The only solution is a=0, b, c any integers.

    If n= 0, that gives k= 0. That is a= b= c= 0, which is included in "a= 0, b, c any integers above".

    If n< 0, k^2= -n/3 so that k= \sqrt{-n/3} which is integer only if n is -3 times a square: [tex]n= -3m^2[tex], in which case k= m.

    That is:
    If n\ge 0 or n< 0 is not of the form n= -3m^2 for some integer m, then a=0, b, c any numbers is the only solution.

    If m= -3m^2, then a= 2m, b= c= m.
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  3. #3
    Super Member Bacterius's Avatar
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    Well n > 3, and I know that (maybe it could help) n = ad (with new unknown d of course)
    And, a > 0, b > 0, c > 0 (and d > 1).

    But it looks like I'm screwed, perhaps I need more information ?
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  4. #4
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    Brute Force Approach - General

    Solve for a using the quadratic formula: 2a=b\pm\sqrt{b^2-4n}=c\pm\sqrt{c^2+4n}

    Now isolate b in terms of c: b^2-4n=(c-b\pm\sqrt{c^2+4n})^2

    Solution: b=\frac{c^2\pm c\sqrt{c^2+4n}+4n}{c\pm\sqrt{c^2+4n}}

    Simplify by multiplying by the conjugate: b=\pm\sqrt{c^2+4n}, or equivalently, c=\pm\sqrt{b^2-4n}

    Substituting, 2a=b+c

    A better way of expressing the solution is by finding b,c such that b^2-c^2=4n, and let a=\frac12(b+c)

    For a to be an integer, b+c must be even, so b,c must have the same parity (both even or both odd), so b-c is even also.

    An all-integer answer can thus be gotten: Given n, factor n, and find b,c satisfying b^2-c^2=4n, or \frac{b+c}{2}\frac{b-c}{2}=n. Define a as above.

    This solution has the advantage of working for all n\in\mathbb{Z} -- there are no restraints on n. Given a random n, the number of a,b,c solutions is equivalent to the number of unique ways to factor n=xy, WLOG x\geq y

    Interestingly, a|n necessarily, so dividing through both equations by a, we arrive at the solution:

    Given any integer n, choose a factor of n, a|n, and define b=a+n/a, c=a-n/a. We might have guessed at this solution from the start, but the preceding serves as a formal proof that it is the whole and only solution (aside from the trivial a=b=c=0, which works iff n=0).
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  5. #5
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    Just wanted to point out what may be an error in HallsofIvy's post .... if you divide 2a^2- a(b+c)= 0 by 2, you get that 2a- (b+c)= 0, and NOT a- (b+c)= 0
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  6. #6
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Bingk View Post
    Just wanted to point out what may be an error in HallsofIvy's post .... if you divide 2a^2- a(b+c)= 0 by 2, you get that 2a- (b+c)= 0, and NOT a- (b+c)= 0
    You mean divide by a surely ?
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  7. #7
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    hahahaha ... yeah, by a
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