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Quadratic Expressions, Perfect Square Not (Posted on 2010-01-03) Difficulty: 2 of 5
Prove that there cannot exist any positive integer x, such that each of 2x2 + 1, 3x2 + 1 and 6x2 + 1 is a perfect square.

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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Hints/Tips still nlng way to go | Comment 4 of 6 |

I still do not have the answer  but maybe my hints will cause
someone to get a formal proof.

Using excel and sloan I arrived  to the following recursion formula:

A(n)= B*A(n-1)-A(n-2)

B , A(0)and A(1) are dependenent on the D in Pell's equation :

D=2          B=6      A(0)=0   A(1)=0  ...    EQ2

D=3          B=4    A(0)=1   A(1)=4     ...    EQ3

D=6          B=10     A(0)=0   A(1)=2    ..   EQ6

And the corresponding series are:

0,2,12,70,408       .....e.g  2*70*70+1 is a square of 99  
1,4,15,56,209,780 ....e.g  3*15*15+1 is a square of 26
0,2,20,198,1960   .....e.g  6*20*60+1 is a square of 49

The 1st and the third series  have 0 as an offset and consist of even  numbers only- the second alternately even and odd.

By visual  or computer-aided inspection of N members one can assume  that there is no a common x present in all three of them, but that does not form a proof.

If some one still wants to make such a comparison -the
recursive formulas might be helpful. However the numbers grow very fast and the comparison would be useful ONLY if a counter example exist i.e. the assumption is wrong.

Still, the way to prove it is not yet clear to me.

 


  Posted by Ady TZIDON on 2010-01-10 13:48:02
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