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An interesting Math question

 
 
vinsan
 
Reply Sun 11 Jan, 2009 01:40 am
Hi Guys,

Just came across this really good math riddle.... can you guys give a hint to solve it using any tradtional mathemtical theorems ?

The sum is like this

Can you find 4 variables a,b,c,d such that

a^5+b^5=c^5+d^5

where ^ indicates raised to power

and a,b,c,d are all unequal positive integers (so nonzero).

(I had written a MATLAB code but really need traditional approach... )

Cheers,
--- Vinit.
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Type: Discussion • Score: 2 • Views: 2,965 • Replies: 7
No top replies

 
BarbieQPickle
 
  1  
Reply Sun 11 Jan, 2009 08:40 pm
@vinsan,
Can all the numbers be raised to different powers?
0 Replies
 
BarbieQPickle
 
  1  
Reply Sun 11 Jan, 2009 08:43 pm
@vinsan,
or is it 5 the power they are raised by?
0 Replies
 
markr
 
  2  
Reply Sun 11 Jan, 2009 08:52 pm
@vinsan,
There are no known solutions. I found this:
Quote:
No solutions to the 5.2.2 equation are known, despite the fact that sums up to 1.02x10^26 have been checked (Guy 1994, p. 140).

at:
http://mathworld.wolfram.com/DiophantineEquation5thPowers.html
0 Replies
 
BarbieQPickle
 
  0  
Reply Sun 11 Jan, 2009 08:54 pm
@vinsan,
Aren't you supposed to add like terms?

4^5 + 6^5 = 7^5 + 3^5

Because:

4^5 + 6^5= 10^5

and

7^5 + 3^5 = 10^5

so

10^5 = 10^5
AttemptATry
 
  1  
Reply Tue 3 Mar, 2009 03:14 pm
@vinsan,
there is actually a very simple solution to this problem.
We have: a^5+b^5 = c^5+d^5 (1)
Therefore: a^5+b^5 - (c^5 + d^5) = 0
We can see that this has the form of w + x - (y + z) = 0 (2)
Now let us take a simple solution that satisfies this equation e.g:
1 + 6 - (3 + 4) = 0 => 7 - 7 = 0

but a^5 = 1 therefore a = 1^(1/5) = 1
b^5 = 6 therefore b = 6^(1/5) = 1.431.....
c^5 = 3 therefore c = 3^(1/5) = 1.246......
d^5 = 4 therefore d = 4^(1/5) = 1.32......

These are positive solutions for a, b, c, d. Choose any values w,x,y,z for (2) that satisfy the equation and you can find an ifinite number of values for a,b,c,d in (1) as w,x,y,z are only bound by w+x = y+z.
rydinearth
 
  1  
Reply Tue 3 Mar, 2009 03:19 pm
@BarbieQPickle,
Quote:
4^5 + 6^5= 10^5

and

7^5 + 3^5 = 10^5


This is not correct.
A^y + B^y does NOT equal (A+B)^y.
As a simple example, try 2^2 + 3^2.
2 squared is 4.
3 squared is 9.
4 + 9 = 13.
(2+3) squared = 5 squared = 25. (not 13).
0 Replies
 
markr
 
  1  
Reply Thu 5 Mar, 2009 02:23 pm
@AttemptATry,
AttemptATry:

Did you not see this:
and a,b,c,d are all unequal positive integers (so nonzero).
0 Replies
 
 

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