This question was inspired by the following question.


Here, we are not looking to factor an integer $N$. We are looking for solutions $n$ to the equation $n^2=pq+y^2$. It turns out that the solutions are many for a given $n$ and are easy to calculate. We need to use the following geometrical figure to help find $p,q,y$ that add up to $n^2$ according to the previous equation.

$$p$$ $$y\phantom{----}y$$ $$q$$

Then we multiply $p$ with $q$ and do the same with the $y's$ and add the result to get $n^2$.
At this point it's useful to choose an integer $n$ to show how the method works. Let $n=10$. We can write $n^2$ as $n^2=10\cdot10=4\cdot25=2\cdot50=5\cdot20$. At this point, we only consider $n^2=10\cdot10$. The factors of $n$ must have the same parity. We see that $n+n=20$. The solutions will use the following values for the pair $(p,q)$.

At this point we don't know which value of $y$ to choose. It turns out that we must have $y+p+y=q$ with the assumption that $q>p$, (by the way, this way to do the multiplications can be used even when $(y,y)$ is replaced by $(y,z)$ except in this case the result will not be $n^2$).

The first pair $(1,19)$ will provide a $y=9$. So we do the multiplication to get $1\cdot19+9^2=10^2$. The pair $(7,13)$ will provide $7\cdot13+3^2=10^2$.

To go through every pair we need to increase $p$ by $1$, decrease $q$ by $1$ to keep the same value for $p+q=20$. $(p,q)$ are not necessarily primes. The values for $y^2$ are $9^2,8^2,7^2,6^2,5^2,4^2,3^2,2^2,1^2$ given in the same order as the pairs $(p,q)$.

As stated in the link given above, "clearly, given a set of $y,p,q$, then $n^2−y^2=pq$ and it follows that $n=p+y$ and $n=q−y$. Hence, $2n=p+q$." Only pairs with both $(p,q)$ primes will contribute to the Goldbach conjecture.

I cannot prove that for a given $n$, we will always have pairs with both $(p,q)$ primes.

Does finding the solution to $n^2=pq+y^2$ help in any way with the Goldbach conjecture?

  • 1
    $\begingroup$ See this similar question. $\endgroup$ Apr 14 '19 at 18:51
  • $\begingroup$ @KeithBackman, thanks for the link. The author did not provide a way to get the solutions to the $n^2=pq+y^2$. . $\endgroup$
    – user25406
    Apr 14 '19 at 19:01
  • 1
    $\begingroup$ surround the dashes with the \phantom{} command and you don't have to talk about them. $\endgroup$
    – user645636
    Apr 15 '19 at 11:46
  • 1
    $\begingroup$ if $n^2=10^2$, $n=10$ not $10\cdot 10$. I uaed the \cdot command for multiplication. $\endgroup$
    – user645636
    Apr 15 '19 at 13:48
  • $\begingroup$ thanks, corrected. $\endgroup$
    – user25406
    Apr 15 '19 at 14:31

The real problem with most algebraic manipulation, is it doesn't matter if p and q are prime. This means we have no real restriction on where to look, this manipulation is similar. Here are some other manipulations:

  • $2n=p+q$

  • $n+n=p+q$

  • $n-p=q-n=d$

  • $n^2-d^2=pq$

  • $p^2=n^2-2dn+d^2$

  • $q^2=n^2+2dn+d^2$

  • $\frac{p^2+q^2}{2}=n^2+d^2$

  • etc.

You can use p and q being odd primes, and properties of n to basically make some of these into case statements. Ex.

$$\begin{cases}p\equiv -q\bmod 6\qquad,\text{if } n\equiv 0 \bmod 6\\1\equiv p\equiv q\bmod 6\qquad,\text{if } n\equiv 1 \bmod 6\\5\equiv p\equiv q \bmod 6\qquad,\text{if } n\equiv 2 \bmod 6\\p\equiv -q\bmod 6\qquad,\text{if } n\equiv 3 \bmod 6\\1\equiv p \equiv q\bmod 6\qquad,\text{if } n\equiv 4 \bmod 6\\5\equiv p\equiv q \bmod 6\qquad,\text{if } n\equiv 5 \bmod 6\end{cases}$$

We also have that neither p or q can be prime divisors of n, because that breaks the distributive property and coprimality, when done with primes. In some cases $2n-3$ can be prime as well. If we could show more primes of certain types than gaps before it in certain arithmetic progressions, we could pigeonhole principle Goldbach's conjecture. But it's not that simple. There's never likely to be a simple algebraic manipulation, that will show it for all cases.

The other complexity is, almost any statement in math, has infinitely many manipulated forms. One of Fermat's little theorem, for example is the following: $$a^{k}\equiv a^{(p-1)x+k} \bmod p $$ gcd(p,a)=1; p a prime;and x any integer. This has infinitely many substitutions possible, including simple algebraic variable, and function substitutions.

  • $\begingroup$ I thought this problem is related to the distribution of semi-primes. I don't known if we have methods to calculate that distribution and how different it will be from the primes distribution itself. $\endgroup$
    – user25406
    Apr 16 '19 at 13:26
  • $\begingroup$ odd distinct semiprimes are the difference of two squares, but so are all numbers that are products of 2 same parity arguments. $\endgroup$
    – user645636
    Apr 16 '19 at 14:05
  • $\begingroup$ Is the information you provided not enough to conclude that there are enough semi-primes to conclude that Goldbach conjecture is true? $\endgroup$
    – user25406
    Apr 16 '19 at 14:10
  • $\begingroup$ which version talks about semiprime density ? $\endgroup$
    – user645636
    Apr 16 '19 at 14:16
  • 1
    $\begingroup$ there's inherently a maximum on the number of pairs, and ir's reached at least once $\pi(n)-\omega(2n)$ $\endgroup$
    – user645636
    Apr 16 '19 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.