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 2021-05-08, 00:09 #1 Charles Kusniec     Aug 2020 Brasil 610 Posts A pretty good challenge A PGC (pretty good challenge). Given: x=(y^2-(n^2-n))/(2n-1)=(y^2-Oblong)/Odd To prove, 1. If n>0, there will be Integer solution for x iff (2n-1) 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). Attached Thumbnails
2021-05-08, 02:18   #2
Dr Sardonicus

Feb 2017
Nowhere

4,673 Posts

Quote:
 Originally Posted by Charles Kusniec A PGC (pretty good challenge). Given: x=(y^2-(n^2-n))/(2n-1)=(y^2-Oblong)/Odd To prove, 1. If n>0, there will be Integer solution for x iff (2n-1) 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).
(y^2 - (n^2 - n))/(2*n - 1) is an integer when

n^2 - n == y^2 (mod 2*n-1)

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 2021-05-08 at 02:21 Reason: Add justification for multiplying congruence by 4

 2021-05-08, 10:38 #3 Charles Kusniec     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.
2021-05-08, 13:37   #4
Dr Sardonicus

Feb 2017
Nowhere

4,673 Posts

Quote:
 Originally Posted by Charles Kusniec Now, the challenge is to find at least one other form of proof where you use the sum of two squares theorem.
I forgot to point out that the thing is also trivial when n = 1, y^2 = 0 (mod 1).

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 two-squares theorem. Using Fermat's "little theorem,"

p divides (2*y)^(p-1) - 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 p-1 is divisible by 4.

Last fiddled with by Dr Sardonicus on 2021-05-08 at 13:38 Reason: Omit unnecessary words!

 2021-05-08, 13:46 #5 Charles Kusniec     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.
 2021-05-08, 15:52 #6 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 80F16 Posts Nice. A numeric example for 2n-1 = 65: https://www.wolframalpha.com/input/?...er+the+integer ETA: No solutions for 2n-1 = 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 2021-05-08 at 16:44
 2021-05-08, 16:50 #7 Charles Kusniec     Aug 2020 Brasil 2×3 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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