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Prime and Twice Square (Posted on 2016-10-22) Difficulty: 4 of 5
Determine all possible values of a prime number P such that each of P+1 and P2 + 1 is twice a perfect square.

Prove that these are the only ones.

No Solution Yet Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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Possible solution. Comment 4 of 4 |

Consider some smaller number, x, such that P = 2x^2-1, and P^2+1 = (2x^2-1)^2, when  (2x^2-1)^2+1= 2y^2.

Then 4x^4-4x^2-2y^2+2 = 0. We can substitute X=x^2 to obtain 4X^2-4X-2y^2+2=0, when X is in {1,4,21,120,687,...}A046090 in Sloane and y is in {1,5,29,169,985,...} A001653 in Sloane. If we can now show that only 1 and 4 are perfect squares for X, we are done, with P=7 as the sole solution.

With modern technology, the proof isn't particularly hard, only tedious. A046090 is computed to a thousand terms in Sloane, some of which are very large. As it is a recurrence relation, it is periodic mod any integer, which makes it possible to compute whether each term is a quadratic non-residue mod a number, say N.

There is a list of quadratic residues in Wikipedia. If we take A046090 under the list of quadratic residues, we find that non-residue solutions (cases where A046090(n) cannot be square) fill up very quickly. For example, N=11, with period 12, has non-residues for n=12k+3,4,8,9,10,11; 19, with period 20, has non-residues for  n=20k+3,5,6,8,13,14,17,18; and 23, with period 11, has non-residues for  n=11k+3,4,5,7,8,10,14,15,16,18,19, producing a table of forms, something like this:

3 4k+3 12k+3 10k+3 20k+3 14k+3 
4 12k+4 5k+4    
5 20k+5 14k+5    
6 8k+6 20k+6 12k+6   
7 4k+3 8k+7 10k+7 12k+7  
8 12k+8 20k+8    
9 12k+9  10k+9 5k+4  
10 12k+10 14k+10    
11 4k+3 12k+11    
12 14k+12     
13 10k+3 20k+13    
14 8k+6 20k+14 5k+4   
15 4k+3 8k+7 12k+3   
16 12k+4     
17 10k+7 20k+17 14k+3   
18 20k+18 12k+6    
19 4k+3 10k+9 12k+7 14k+5 5k+4 
20 12k+8     
21 12k+9     
22 8k+6 12k+10    
23 4k+3 12k+11 8k+7 10k+3  
24 14k+10 5k+4    
25 20k+5     
26 20k+6 14k+12    
27 4k+3 10k+7 12k+3   
28 20k+8 12k+4    
29 10k+9 5k+4    
30 8k+6 12k+6    
31 4k+3 8k+7 12k+7 14k+3  
32 12k+8     
33 10k+3 20k+13 12k+9 14k+5  
34 20k+14 5k+4 12k+10   
35 4k+3 12k+11    
36 11k+3     
37 10k+7 20k+17    
38 20k+18 14k+10    
39 4k+3 10k+9 5k+4   
40 14k+12

etc. There are probably neater ways of completing the table using ideal candidates, i.e. those with short periods and a low common multiple (e.g, 41 has a period of only 6, taken with A046090).

Hence n=1,2, implying that A046090(n) =1, 4 are indeed the only squares in A046090, and P=7 is the sole solution to the puzzle.

In fact, I've suspected for a while that there is a large class of these sequences exhibiting the same property - only two square terms; e.g. A001653. I suspect there is a more elegant and more generalised solution.


Edited on October 24, 2016, 1:15 am
  Posted by broll on 2016-10-23 22:37:56

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