# Prime number conjecture

It was suggested that I put my full conjecture up instead of specific examples. Here it is:

Given any prime p>3, there exists c such that the following conditions hold:

1a. The quadratic equation $x^2-px-c=0$ has integer solutions

2a. c is divisible by every prime less than $\sqrt{p}$

3a. c is not divisible by any other prime

Note that this is equivalent to:

There exists integers A and B such that:

1b. $A\pm B=p$

2b. $A\times B=c$, which follows conditions 2a and 3a above.

And lastly (for now) this is equivalent to:

If $X_T={X(X+1) \over 2}$ = the $X^{th}$ triangular number (1+2+3+...+X), which is also equal to $_{X+1}C_2$, then

given prime number p, and $n={p-1\over 2}$

There exists positive integers m and c such that:

$2(m_T-n_T)=c$, and c follows conditions 2a and 3a above.

As hinted before, this conjecture holds for every prime up to 397; my computer goes too slow at that point to tell if it continues to hold. It may be that it will hold for some humongous powers of the factors of c. It is essentially a question of how high is high enough to conclude it disproven, but we can never know empirically since we can't try every possible exponent. We just have to find a proof. I will try to find a way to post my PARI program without taking too much space here. Thanks for your continued support!

• Is (1a, 2a, 3a) a correct statement of your conjecture? The phrase "lastly for now" suggests some condition added on. – Will Jagy Oct 30 '14 at 23:32
• Instead of the program, how about listing $(p,c)$ and the full factorization of $c$ for $p < 20?$ – Will Jagy Oct 30 '14 at 23:40
• In other words, the conjecture is that for any $p$, we can always find an integer $A$ such that both $A$ and $(A+p)$ are only divisible by primes smaller than $\sqrt{p}$, and moreover every such prime is part of the factorization of $A(A+p)$? (And the quadratic is $(x+A)(x-(A+p))=x^2-px-A(A+p)$.) – alex.jordan Oct 31 '14 at 0:54
• By "lastly for now" I simply meant we have 3 different equivalent conjectures but there may be more different ways of expressing the conjecture in the future. – Aaron Horak Oct 31 '14 at 12:13
• If I had time I would supply more, but here are the numbers that Will Jagy asked for: 5=4+1; 7=8-1; after 9 they must have factors of 2 and 3, until we hit 25, so, 11=9+2=27-16; 13=9+4=16-3; 17=9+8=81-64; and 19=16+3=27-8 – Aaron Horak Oct 31 '14 at 21:00

The conjecture cannot be true, if 1a is included, because for that quadratic to hold, $A+B = P$ and the largest possible value of $AB$ is $\frac{p^2}{4}$ so for sufficiently large primes $AB$ cannot include factors of all the primes smaller than $\sqrt{p}$.

So let's change the conjecture to:

For any prime $p>3$ there exist integers $a, b$ such that $a-b = p$ or $p = a+b$ and $ab$ is divisible by all primes less than $\sqrt{p}$ and by no other primes. You need the $a+b$ possibility to cover small primes like $5$.

Considerations about the density of primes certainly suggest that this is true for all sufficiently large primes, but of course that is not a proof.

• I think you are assuming $A$, $B$ are the same sign. But $|AB|$ can be larger than $\frac{p^2}{4}$ if one of $A$, $B$ is negative and the other positive. – alex.jordan Oct 31 '14 at 0:50
• Yes, exactly. 4+1=5 and the quadratic would be $x^2-5x+4$ – Aaron Horak Oct 31 '14 at 12:16
• posted other solutions for primes, see above, will try to add more monday – Aaron Horak Oct 31 '14 at 21:02
• 179=1089-910;181=286-105;191=455-264;193=1183- 990;197=1352-1155;199=364-165;211=715- 504;223=1848-1625;227=3087-2860;229=385- 156;233=728-495;239=330-91;241=780-539;251=1001- 750;257=455-198;263=770-507;269=819-550;271=546- 275;277=420-143;281=1001-720;283=1053- 770;293=2295-2002;307=2925-2618;311=20111- 19800;313=18513-18200;317=11900-11583;331=2431- 2100;337=1724800-1724463;347=7497-7150;349=910- 561;353=3213-2860;359=1430-1071;367=7150- 6783;373=88825-88452;379=35700-35321;383=53865- 53482;389=3315-2926 – Aaron Horak Nov 3 '14 at 18:29
• Here (above) I have listed the last results output by my program. If you factor the numbers you will get a better understanding of how this all works. The trouble with an infinite number of possible exponents is illustrated well by 337, and then the next few have much smaller exponents. – Aaron Horak Nov 3 '14 at 18:29