20210508, 00:09  #1 
Aug 2020
Brasil
2·3 Posts 
A pretty good challenge
A PGC (pretty good challenge). Given:
x=(y^2(n^2n))/(2n1)=(y^2Oblong)/Odd To prove, 1. If n>0, there will be Integer solution for x iff (2n1) are the numbers that are divisible only by primes congruent to 1 mod 4 (http://oeis.org/A004613, https://oeis.org/A008846, https://oeis.org/A020882). 2. If n≤0, there will be Integer solution for x iff n=2m negative Even (negative sequence https://oeis.org/A226485, or twice the negative sequence http://oeis.org/A094178). 
20210508, 02:18  #2  
Feb 2017
Nowhere
2^{4}×7×43 Posts 
Quote:
n^2  n == y^2 (mod 2*n1) 4*n^2  4*n == 4*y^2 (mod 2*n  1) ^{†} (2*n  1)^2 == 4*y^2 +1 (mod 2*n  1) 4*y^2 + 1 == 0 (mod 2*n  1); that is, 2*n  1 divides 4*y^2 + 1 Since 4*y^2 + 1 is divisible only by primes congruent to 1 (mod 4), the same is true of its divisor 2*n  1. This solves (1) straightaway. If n = 0, we have y^2 == 0 (mod 1) which is trivial. If n < 0, we have that 2*n + 1 divides 4*y^2 + 1. This implies all divisors of 2*n + 1 are congruent to 1 (mod 4), so that n is even, solving (2) in the formulation of the second OEIS sequence. ^{†}Multiplying the congruence through by 4 is allowed, since 4 is relatively prime to 2*n  1. Last fiddled with by Dr Sardonicus on 20210508 at 02:21 Reason: Add justification for multiplying congruence by 4 

20210508, 10:38  #3 
Aug 2020
Brasil
2×3 Posts 
Dear Dr Sardonicus, I read your proof and found no fault. Apparently it is perfect. Congratulations. I think the way I asked already giving the answers helped. Now, the challenge is to find at least one other form of proof where you use the sum of two squares theorem. A very interesting surprise will appear, and that is the only hint for now.

20210508, 13:37  #4  
Feb 2017
Nowhere
2^{4}·7·43 Posts 
Quote:
What I proved was that your given hypotheses imply that 2*n  1 divides 4*y^2 + 1 if n > 1 2*n + 1 divides 4*y^2 + 1, if n < 0. The fact that every prime divisor p of 4*y^2 + 1 is congruent to 1 (mod 4) is elementary, and does not require the twosquares theorem. Using Fermat's "little theorem," p divides (2*y)^(p1)  1. From the above, p divides (2*y)^4  1, but p does not divide (2*y)^2  1. Therefore, 2*y has multiplicative order 4 (mod p), from which it follows that p1 is divisible by 4. Last fiddled with by Dr Sardonicus on 20210508 at 13:38 Reason: Omit unnecessary words! 

20210508, 13:46  #5 
Aug 2020
Brasil
2·3 Posts 
Actually, we have 2 trivialities for n. We have n=0 and n=1 as trivialities. I agree with what you said. What I am complementing is that there is another way to do this proof and in it we have to use the sum of 2 squares theorem. When we do the proof by this second way, we find a very interesting equality.

20210508, 15:52  #6 
"Rashid Naimi"
Oct 2015
Remote to Here/There
3·5·139 Posts 
Nice.
A numeric example for 2n1 = 65: https://www.wolframalpha.com/input/?...er+the+integer ETA: No solutions for 2n1 = 21 even though it is of the form 4m+1 since it has at least one prime factor of the form 2m+1 with odd m: https://www.wolframalpha.com/input/?...er+the+integer Last fiddled with by a1call on 20210508 at 16:44 
20210508, 16:50  #7 
Aug 2020
Brasil
6_{10} Posts 
Dear a1call,
Thanks for the example. I like you used 65 as an example, because this is one of those cases of multiplicity of primitive Pythagorean triangle. At 65, we have 2 solutions because we have 2 ways to make the Pythagorean triangle. That’s why I don’t really like how WolframAlpha and many others present this kind of solution. The way they present it looks like we have 4 sequences of solutions, when in fact, we only have 2 sequences. They should present only the sequences a(n) = 65 n^2 + 8 n  16 and a(n) = 65 n^2 + 18 n  15. These two sequences are at offset Zero. For more details about offset I have a study that explains well this phenomenon... 
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