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Finding Integer Solutions to Four-Variable Equation

Date: 06/10/2007 at 02:54:08
From: Shamik
Subject: Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d

a, b, c and d are four natural numbers such that a^2 + b^2 + c^2 + 
d^2 = a*b*c*d.  Find all values of a, b, c and d where all of them are 
different integers.

Obviously, a = b = c = d = 2 is a solution as 2^2 + 2^2 + 2^2 + 2^2 = 
16 = 2*2*2*2.

I tried to use (A^2 + B^2)*(C^2 + D^2) = (AC - BD)^2 + (AD + BC)^2
                                       = (AD - BC)^2 + (AC + BD)^2

I also had a look at http://www.alpertron.com.ar/4SQUARES.HTM 

which says "Every positive integer is a sum of four integer squares." 
Say, N =(a^2 + b^2 + c^2 + d^2)*(A^2 + B^2 + C^2 + D^2) = (aA + bB + 
cC + dD)^2 + (aB - bA + cD - dC)^2 + (aC - bD - cA + dB)^2 + (aD - dA 
+ bC - cB)^2

Now, if I expand this (aA + bB + cC + dD)*(aB - bA + cD - dC)*(aC - 
bD - cA + dB)*(aD - dA + bC - cB), will it be equal to N = (a^2 + b^2 
+ c^2 + d^2)*(A^2 + B^2 + C^2 + D^2) in some special cases?

I am asking this as I am not in a position to continue further. 
Please, set me on the right track with further help if I'm heading 
towards the wrong direction.

I was exploring with the numbers:
 
(i) (1,2,3,4)
S = 1^2 + 2^2 + 3^2 + 4^2 = 30
P = 1*2*3*4 = 24

(ii) (1,2,4,5)
S = 1^2 + 2^2 + 4^2 + 5^2 = 46
P = 1*2*4*5 = 40 

(iii) (1,2,11,12)
S = 1^2 + 2^2 + 11^2 + 12^2 = 270
P = 1*2*11*12 = 264

(iv) (1,2,29,30)
S = 1^2 + 2^2 + 29^2 + 30^2 = 1746
P = 1*2*29*30 = 1740

(v) (1,2,100,101)
S = 1^2 + 2^2 + 100^2 + 101^2 = 20206
P = 1*2*100*101 = 20200

This was the closest I could get where S - P = 6 but I could never 
get S = P.  I understand what I posted above is just a kind of 
experimental data.  It doesn't prove or disprove anything.

Please, help me.  Thanks and regards,

Shamik




Date: 06/11/2007 at 16:43:35
From: Doctor Vogler
Subject: Re: Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d

Hi Shamik,

Thanks for writing to Dr. Math.  The trick to solving this problem is 
noticing that you can get from one solution to another by replacing d 
by a*b*c-d.  In fact, there are three ways to get new integer 
solutions from a known integer solution:

  (1) Change the signs of two of the four variables
  (2) Permute (or reorder) the four variables
  (3) Replace d by a*b*c - d.

Of course, if any variable is zero, then they must all be zero, and 
the all-zero solution is unchanged by all three of the above 
actions.  Action (1) always commutes with actions (2) and (3), and 
it's easy to prove that you must have an even number of positive 
variables, so you can really just about ignore (1).  But (2) and (3) 
are very very important.

For example, we have the solution (2, 2, 2, 2) that you mentioned.  
Using (3), we get the solution (2, 2, 2, 6).  Using (2) and then (3), 
we get the new solution (2, 2, 6, 22).  Using (2) and then (3), we 
can get either of the solutions (2, 2, 22, 82) or (2, 6, 22, 262).  
Here we have a solution with a < b < c < d, so this answers your 
first question that it is possible.  Next, using (2) and (3) on those 
solutions, we can get five more solutions:

  (2, 2, 82, 306)
  (2, 22, 82, 3606)
  (2, 6, 262, 3122)
  (2, 22, 262, 11522)
  (6, 22, 262, 34583).

Then we can use (2) and (3) on those five solutions to get 14 more, 
and so on to infinitely many integer solutions (one at each level 
will have a=b=2, but all of the others will have four different 
values).

Next, you should try to prove that these are all integer solutions.  
To do this, first you should prove that from any integer solution 
(except for the all-zero solution), you can use (1), (2), and (3) a 
finite number of times to get to one solution which satisfies

  0 < a <= b <= c <= d <= (1/2)abc.

The idea is to use (1) to fix the signs, (2) to order the four 
variables, and then (3) to make d smaller than (1/2)abc.  But then 
that might change the order, so you might have to repeat (2) and (3) 
until you get the above inequalities, but since d keeps getting 
smaller at each step, and there are only finitely many small 
integers, this process will eventually stop.

Next, you should prove that there is exactly ONE integer solution 
which satisfies

  0 < a <= b <= c <= d <= (1/2)abc.

Proving this requires a lot of inequalities to prove that a, b, and c
can't be very large, and then you try the small cases and see which
ones work.

And then you're done!  Do you see how proving the above things would 
prove that all solutions come from (2, 2, 2, 2) by using the 
transformations I described?

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/ 




Date: 06/12/2007 at 03:54:34
From: Shamik
Subject: Thank you (Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d)

Docter Vogler,

Thank you very much. It was a new learning for me on how to attempt
these kind of problems.

High regards,

Shamik
Associated Topics:
College Number Theory

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