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Cubic Problem: Existence Of A Solution (Posted on 2005-12-28) Difficulty: 3 of 5
Does there exist any positive integral solution of the equation given below?

16(X³+Y³) + 12(X²+Y²) + 3(X+Y) = 2,138,156,388

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

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Solution Full solution | Comment 6 of 8 |

There are no integer solutions at all !

For the proof, assume the contrary that there are x, y such that

3*(x + y) + 12*(x^2 + y^2) + 16*(x^3 + y^3) = 2138156388.

Then the remainders on each side after dividing each by n will also be the same for ALL integers n. That is

3*(x + y) + 12*(x^2 + y^2) + 16*(x^3 + y^3) = 2138156388. (mod n)

After a bit of trial and error, we let n = 7 and test for x, y = 0 ... 6

Now 2138156388 = 4 but 3*(x + y) + 12*(x^2 + y^2) + 16*(x^3 + y^3) can only equal {0,1,3,5,6}.

Hence the assumption that there are integer solutions is false.

The method also works here for n = 14, 21 ,...

 

 

 


  Posted by goFish on 2005-12-29 07:11:42
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