I've developed an equation that solves a problem I'm working on, but the only way I know how to solve it is by incrementally trying values of $n$ from $1 \to \infty$ until I arrive at the solution. I'm hoping someone can point me to some relevant research/theory that finds solutions in a different way.

The equation is of the following form, where all values are positive integers and $a$ and $b$ are constant:

$$a + 2n(b-n) \equiv 0 \mod (b-2n)$$

For example,

$$74003 + 2n(47595 - n) \equiv 0 \mod (47595 - 2n)$$

is solved when $n = 5456$. The values of $a$ and $b$ are chosen such that a solution always exists before $2n \ge b$. I'm only interested in the smallest $n$ that first satisfies the equation, not all possible solutions. Any suggestions or reformulations of the equation are welcome.

  • 2
    $\begingroup$ Possible strategy. You want to find the smallest $n$ for which there is a $k$ such that $a+2n(b−n) = k(b−2n)$. That is a quadratic equation in $n$. The solution can be an integer only when the discriminant (which depends on $a$, $b$ and $k$) is a perfect square. Perhaps a search for the $k$ that works will be faster than the direct search for $n$. $\endgroup$ – Ethan Bolker May 9 '19 at 23:29
  • $\begingroup$ @EthanBolker Could you elaborate a little? I think I understand what you're saying, but I don't see how I eliminate $n$ when solving for $k$. It seems like I now have 2 variables to deal with. $\endgroup$ – EntangledLoops May 9 '19 at 23:35
  • $\begingroup$ See my suggested answer. $\endgroup$ – Ethan Bolker May 9 '19 at 23:50

I'm adding a new answer instead of updating or appending to my original one since this is quite different based on the new information provided. The answers by Ethan Bolker and Fabio Somenzi involve using

$$a + 2n(b - n) = k(b - 2n) \tag{1}$$

to create a quadratic equation in $n$, i.e.,

$$n^2 - (k + b)n + \frac{bk - a}{2} = 0 \tag{2}$$

which has a discriminant of

$$D = (k + b)^2 - 2(bk - a) = k^2 + b^2 + 2a \tag{3}\label{eq3}$$

The value of $n$ is

$$n = \frac{k + b \pm \sqrt{D}}{2} \tag{4}\label{eq4}$$

The smallest positive integer $n$ comes from subtracting $\sqrt{D}$ where $D$ is a perfect square of the same parity as $k + b$, and less than but as close as possible to $(k + b)^2$. Thus, from \eqref{eq3}, you want to start with the smallest $k$ such that $bk - a \gt 0$, say

$$k_0 = \left\lfloor \frac{a}{b} \right\rfloor + 1 \tag{5}\label{eq5}$$

This gives

$$b \ge bk_0 - a > 0 \; \Rightarrow \; -2b \le -2(bk_0 - a) \lt 0 \tag{6}\label{eq6}$$

Using this in \eqref{eq3},

$$(k_0 + b)^2 - 2(bk_0 - a) \ge k_0^2 + 2bk_0 + b^2 - 2b \tag{7}\label{eq7}$$

However, note that

$$(k_0 + (b - 1))^2 = k_0^2 + 2bk_0 + b^2 - 2b + (1 - 2k_0) \tag{8}\label{eq8}$$

Thus, the next smaller perfect square of $D$ is $(k_0 + (b - 1))^2$. Note going from $m^2$ to $(m + 1)^2$ involves adding $2m + 1$, then to $(m + 2)^2$ involves adding $2m + 3$, then add $2m + 5$ to get $(m + 3)^2$, etc. In other words, even for very large integers, you can quickly and easily go from a perfect square to the next perfect square by adding a value which you increment by $2$ each time. This is generally faster than incrementing a value and then squaring it, as https://agner.org/optimize/optimizing_cpp.pdf says at section 14.4 Integer multiplication:

Integer multiplication takes longer time than addition and subtraction (3 -10 clock cycles, depending on the processor).

For very large integers, using some integer handling package, multiplication can be quite a bit slower. In fact, a very fast method to handle extremely big calculations, in $O(n\log(n))$ for $n$ digits, was just announced recently, e.g., the April $4$, $2019$ article at Mathematicians Develop New Algorithm for Multiplying Large Numbers.

Starting at $k = k_0$ (from \eqref{eq5}) in \eqref{eq3}, and using \eqref{eq8}, you can quickly determine each next value for $k$ and compare it to the next possible perfect square, incrementing each value(s) as appropriate, until the values match, and the result gives an integer in \eqref{eq4}. This is relatively efficient, and is generally considerably faster than doing integer divisions (e.g., for factoring, such as in my original answer, or to check modulo values). For example, https://agner.org/optimize/optimizing_cpp.pdf says at section 14.5 Integer division:

Integer division takes much longer time than addition, subtraction and multiplication (27 - 80 clock cycles for 32-bit integers, depending on the processor).

This speed issue will vary with aspects like the size of the integers, the compiler being used, etc., but I believe in basically all cases integer division will be quite slow.

  • $\begingroup$ Thanks for another detailed answer. I'm still parsing/testing this and will accept it if it works. $\endgroup$ – EntangledLoops May 20 '19 at 0:10
  • $\begingroup$ @EntangledLoops You're welcome. Note I fixed a typo in the square term in the middle part of eq. $3$. Also, I've added a few other details, such as why my suggestion for $2$ adds to get each next perfect square is faster than an increment and multiplying instead. Good luck with your testing and, regardless of whether or not this helped, I would appreciate a brief note re: what you found. Thanks. $\endgroup$ – John Omielan May 20 '19 at 1:39
  • $\begingroup$ It sounds like you've essentially reverse engineered what got me here in the first place: entangledloops.com/files/semiprime/semiprime.pdf (See algorithm 2: Coprime Solver). I'm still checking the math to see if we're ultimately doing the same calculation, or your way of getting to the next square is different. $\endgroup$ – EntangledLoops May 20 '19 at 16:47

The congruence equation to solve for is

$$a + 2n(b-n) \equiv 0 \pmod{(b-2n)} \tag{1}\label{eq1}$$

You can do several manipulations of it to eliminate the $n$ on the left hand side as follows:

\begin{align} a + 2n((b - 2n) + n) & \equiv a + 2n^2 \\ & \equiv 2a + 4n^2 \\ & \equiv 2a + (-2n)^2 \\ & \equiv 2a + ((b - 2n) - b)^2 \\ & \equiv 2a + b^2 \\ & \equiv 0 \pmod{(b-2n)} \tag{2}\label{eq2} \end{align}

Assuming $a$ and $b$ are not too large, there are relatively efficient algorithms to factorize $2a + b^2$ which you can find online (e.g., Stack Overflow's Efficient Prime Factorization for large numbers), use an appropriate library or perhaps your programming language even has a quite good one built in. You can then check the factors to find the largest one which is less than $b$ and has the same parity. There may also be methods which can quite quickly find factors close to a given value without necessarily factorizing the entire value, but a quick search didn't show me anything. Using something like this could be generally faster than checking each value of $n$ in turn, but it depends on things like how efficient your factoring algorithm is, the speed of checking congruences, etc. As such, likely doing some timing testing may be the only way to determine which method is actually usually faster in your particular situation.

One other issue to consider with getting all of the factors of $2a + b^2$ is that, although you're only interested in the smallest $n$ now, if you ever find you would like to know all such $n$, then for each of the factors $f$ less than $b$ with the same parity, you get $n = \frac{b - f}{2}$.

Update: Based on the feedback of the OP in the comments re: "$a$ and $b$ can grow to be arbitrarily large ...", an alternative which may work better is to use a loop to check either $a + 2n^2$ or $2a + b^2$ for each $b - 2n$ value instead of the original $a + 2n(b - n)$. However, I'm not sure how much, if any, this will improve the speed, but my suspicion is that it won't be very much. As I stated earlier, it may be difficult to tell what will work better by analysis and, instead, testing various methods may be the only way to determine, with relatively strong confidence, which method works best for solving this problem.

  • $\begingroup$ Thanks for your answer. Unfortunately, $a$ and $b$ can grow arbitrarily large in this case, as they are related to large semiprime numbers. I'll take a look at the provided link and see if I can apply any of those techniques, regardless. $\endgroup$ – EntangledLoops May 10 '19 at 4:44
  • $\begingroup$ @EntangledLoops You are welcome, & thanks for your extra info. I assume your current algorithm basically calculates $2n(b - n)$, adds it to $a$ & then checks if $b - 2n$ divides evenly into it. If so, you may wish to at least check $a + 2n^2$ instead as it could be a bit more efficient. Alternatively, although the value may be considerably larger, it may be faster to check if each next $b - 2n$ divides into $2a + b^2$ instead. Finally, note if $2a + b^2$ is too large to efficiently factor, then it might also be difficult to efficiently check Ethan's idea of $k^2 -6bk +4a + b^2$ being a ... $\endgroup$ – John Omielan May 10 '19 at 5:24
  • $\begingroup$ @EntangledLoops (cont.) perfect square for each $k$. This is something I don't know, but to me at least, it's not clear what way may be best, and I suggest you perhaps implement & try the various methods to see what works best for you. Note I'll update my answer with my suggestions in my comment above for using the algorithm that I wrote about. $\endgroup$ – John Omielan May 10 '19 at 5:27
  • $\begingroup$ I implemented your idea of computing $2a+b^2$ and then checking every $b-2n$ until a solution is found. It works, but differs by only a few milliseconds from my current implementation for various sizes of $a$, $b$. It seems I need a smarter technique then just trying values of $n$, or at least need to narrow the bounds on $n$ first. $\endgroup$ – EntangledLoops May 10 '19 at 23:39

Possible strategy.

You want to find the smallest $n$ for which there is a $k$ such that $$ a+2n(b−n)=k(b−2n). $$ That is equivalent to $$\ n^2 + (b-k)n + bk-a = 0 $$ (if I did the algebra correctly).

When you solve that equation for $n$ as a function of $a$, $b$ and $k$ the answer will be an integer only if the discriminant $$ (b-k)^2 - 4(bk-a) = k^2 -6bk +4a + b^2 > 0 $$ is a perfect square (if I did the algebra correctly).

It may be easier to test that with values of $k$ until you get a square than to test for $n$ in the original form of the problem.

  • $\begingroup$ Thanks, this is in line with what I'm looking for. I'll give the determinant idea a try and accept this answer if nobody else has a better suggestion. $\endgroup$ – EntangledLoops May 10 '19 at 4:47
  • $\begingroup$ @EthanBolker Expanding $a+2n(b−n)=k(b−2n)$ gives $a + 2bn - 2n^2 = kb - 2kn$. Moving everything to the right gives $0 = 2n^2 - 2kn - 2bn + kb - a = 2n^2 - 2(k + b)n + kb - a$. The discriminant would then be $(2(k+b))^2 - 4(2)(kb - a) = 4(k^2 + 2a + b^2)$. Interestingly, the $2a + b^2$ term is what I came up with in my answer. Also, for checking, you would want to start with the smallest $k$ such that $kb - a \gt 0$, and then increase $k$, to give the smallest resulting positive value of $n$. $\endgroup$ – John Omielan May 10 '19 at 6:49

This answer should be a comment to @EthanBolker's answer, but is a bit too long for that.

From $a+2n(b-n) = k(b-2n)$ we get

$$ n^2 -(k+b)n + \frac{bk-a}{2} = 0 \enspace. $$

The discriminant of this quadratic is

$$ k^2 + b^2 + 2a \enspace. $$

If $b$ is odd, then $k = \frac{b^2 + 2a - 1}{2}$ makes the discriminant a perfect square, but this is the largest value of $k$, not the smallest.

Suppose then $k^2 + b^2 + 2a = (k+\delta)^2$. Then

$$ \delta(2k+\delta) = b^2 + 2a \enspace. $$

The smallest $k$ for which $k^2 + b^2 +2a$ is a perfect square is obtained when $\delta$ is a factor of $b^2 + 2a$ close to the square root of $b^2 + 2a$. In a way, we have connected to the observation by @JohnOmielan about the usefulness of factoring $b^2 + 2a$.

  • $\begingroup$ This is an interesting answer. However, note the OP wants the smallest $n$, not the smallest $k$. As I commented to Ethan's answer, this comes from the smallest $k$ such that $kb - a \gt 0$ and which satisfies the discriminant to give an integral $n$. This extra condition helps give an additional connection with, as you stated, my answer about the potential usefulness of factoring $2a + b^2$. $\endgroup$ – John Omielan May 10 '19 at 7:19
  • $\begingroup$ @JohnOmielan Thanks for pointing out that the objective is minimum $n$. I had lost sight of it. $\endgroup$ – Fabio Somenzi May 10 '19 at 7:58
  • $\begingroup$ Unfortunately, $b^2 + 2a$ turns out to be the original problem I'm trying to simplify. My current algorithm derives $a$ and $b$ from $s$, a large semiprime number. I've found a way to use modular arithmetic to find a congruency with a much smaller number that will also be congruent to $s$ (i.e. a smaller number coprime with $s$), which directly gives me the factors. For this, I'm using $(a+b) \mod (a-2n)$, and I'm trying to understand why this works so I can improve it. $\endgroup$ – EntangledLoops May 11 '19 at 0:05
  • $\begingroup$ I just realized why this works. It's because I've computed $s = b^2 + 2a$, and then am searching for numbers congruent to $b+a$, which will also be congruent to $s$. $\endgroup$ – EntangledLoops May 11 '19 at 0:14
  • $\begingroup$ @EntangledLoops I've added a new answer giving what I believe is a more efficient way to solve the problem. Note I've tried several things, but the first $2$ equations show properly when I'm in edit mode, but not when they're displayed afterwards. I found the same issue occurred with a different browser on a different machine. My best guess is that it's an issue with something showing above it, so the problem may go away depending on the display order & any other text in between. $\endgroup$ – John Omielan May 12 '19 at 8:49

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