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Four Variable Diophantine Expression

Date: 05/10/2008 at 02:50:38
From: Abhi
Subject: trying to find number of digits in square of the number

For what pairs of different positive integers is the value a/(a+1) +
b/(b+1) + c/(c+1) + d/(d+1) an integer?  How would I solve it?



Date: 05/10/2008 at 14:20:59
From: Doctor Vogler
Subject: Re: trying to find number of digits in square of the number

Hi Abhi,

Thanks for writing to Dr. Math.  I would recommend the following:
 
First assume that a < b < c < d, because if you can find all solutions
that satisfy this inequality, then you can find all solutions by
rearranging the variables.

Next, since your variables are positive integers, and since you can
try small positive integers, you should always ask why the equation
can't be true when the variables are *large* positive integers.  Well,
if they are, then each fraction will be slightly less than 1, which
means that the sum will be something between 3 and 4.  Indeed, if each
variable is bigger than 3, then each fraction will be between 3/4 and
1, and therefore the sum will be between 3 and 4.  Can you prove that?
In fact, no matter what a, b, c, and d are, the sum can't be 4 or more.

Therefore, a can't be bigger than 3.  It also can't equal 3, since
then b, c, and d would have to be bigger than 3, and the sum would
still be between 3 and 4.  Therefore, a = 1 or a = 2.

If a = 2, then b >= 3 and c >= 4 and d >= 5, so the sum will be at
least 2/3 + 3/4 + 4/5 + 5/6 > 3.  So again the sum is between 3 and 4.
Therefore, a = 1.

Even if a = 1, the sum will be at least 1/2 + 2/3 + 3/4 + 4/5 > 2, so
the only way we could get an integer is if we get 3.  Also, if a = 1,
and the sum equals 3, then the last three fractions have to add up to
5/2.  So that won't happen if each of them is bigger than 5/6, which
means that b can't be 5 or more.  Therefore, b is between 2 and 4
inclusive.

If b = 4, then c/(c+1) + d/(d+1) = 17/10, which means that c/(c+1),
which is smaller than d/(d+1), must be less than 17/20 < 6/7.  So c
has to be 5.  But then d/(d+1) = 17/10 - 6/7 = 59/70, which is impossible.

If b = 3, then c/(c+1) + d/(d+1) = 7/4, which means that c/(c+1) must
be less than 7/8.  So c has to be between 4 and 6 inclusive.  Do any
of those numbers give a valid choice for d?

Finally, if b = 2, then c/(c+1) + d/(d+1) = 11/6, which means that
c/(c+1) must be less than 11/12.  So c has to be between 3 and 11
inclusive.  Do any of those numbers give a valid choice for d?

Those are the only possibilities, except for rearranging the numbers
in the solutions you find.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.


- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory
High School Number Theory

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