No unique name has been given to this equation.

It is solvable over integers.

Method of solving is given in the attached PDF file.

This is the equation of a conic, as can be seen, for instance, as follows:

Set (x+y)/2 = u and (x-y)/2 = v, thus x = u+v and y = u-v.

The equation then takes the form

u2-v2+a(u+v)+b(u-v) = c (u+(a+b)/2)2-(v-(a-b)/2)2 = c + ab

(2u+a+b)2-(2v-(a-b))2=4(c+ab)

Now set 2u+a+b=U, 2v-a+b=V and 4(c+ab)=R2 to find

U2-V2=R2

(If c+ab

Hi Stam Nicolis, Thank you very much. It is really a great help to me.

Shouxing Qu You're welcome.

Hi all.

Regarding the discussion, I would like to note the following.

The factorization in the first answer leads to an equation that is equivalent to the initial equation; i.e. the two equations have the same solutions; not only in integers, but in reals too. This is trivial to see; the two equations differ only in the addition and subtraction of ab. Then there exist two cases about ab+c; i.e. ab+c being either zero or non-zero; Next, if ab+c is zero, the solutions are given trivially; if ab+c is non-zero, the simple observation that x+b is a divisor of ab+c leads again to all integer solutions. Nothing is missing. Of course, the parametrization of the solution can change, but the set of solutions does not change.

What matters is that, if ab+c is a prime, p, then, either x+b = 1 and y+a = p, or x+b=p and y+a=1, so there are two solutions; while, if ab+c isn't a prime, then there is a finite number of solutions. However counting them-and finding them-isn't as easy as it might seem from the examples provided, because, if ab+c = d1d2...dn, with n>2, then there are many ways of partitioning this product into two factors, that will then can be identified with x+b and y+a respectively and it's necessary to find a useful way of counting them. That's where the rational parametrization of the hyperbola is useful.

For instance if ab+c = d1d2d3, with three distinct factors, then the cases are

x+b = d1, y+a = d2d3

x+b = d2, y+a = d1d3

x+b = d3, y+a = d1d2

along with

y+a = d1, x+b = d2d3

y+a = d2, x+b = d1d3

y+a = d3, x+b = d1d2

giving six solutions.

Now, of course, it's possible to write

(2u+a+b)2-(2v-(a-b))2=4(x+b)(y+a)=4(c+ab);

so the expression that appears in @Mahalakshmi Marikani's solution, appears in this approach, too-a useful sanity check!; the problem remains counting the solutions. This is where using the rational points on the hyperbola helps.

What remains to be understood is that @Mahalakshmi Marikani's solution implies that a solution always exists, whereas the correspondence with the integer points on the hyperbola seems to exist only when ab+c is the square of an integer...

On the other hand, one might argue that it's not necessary for 4(c+ab) to be a square of an integer; it's just necessary to find the integer points on the hyperbola

U2-V2=4(c+ab)

which can exist, also, when 4(c+ab) isn't the square of an integer; it suffices that it is an integer.

So there are two cases: (a) When 4(c+ab)=R2, with R integer; and (b), when 4(c+ab) is not the square of an integer.

Given any positive integer n, by the fundamental theorem of arithmetic, there exist positive integers m, n1, n2,…,nm, which are unique, and prime numbers p1,…,pm, which are also unique, such that n=p1n1…pmnm.

For every i=1,…,m, the factor pini gives the positive divisors, 1,pi,pi2,…,pini; whence it gives ni+1 divisors. Then, by the product rule for counting, the number of positive divisors of n are (n1+1)… (nm+1). Actually, this is the well-known number-of-divisors function. Now, in order to list all positive divisors of n, one can write all different divisors coming from the factor pini, for every i=1,…,m, which are 1,p1,…,p1n1,p2,…,p2n2,…,pm,…,pmnm; they are n1+n2+…+nm+1 in total, and then taking all possible combinations of 1,2,…, n1+n2+…+nm+1 elements from the previous sequence, one obtains each positive divisor only one time (no repetitions of the same divisor). Clearly, by the positive divisors, one can trivially find the negative divisors, too. Then, putting each divisor in place of di in M. Mahalakshmi 's solution for the non-trivial case of ab+c being non-zero, one obtains all integer solutions and each solution is obtained one time (no repetitions of the same solution), as one can easily see by M. Mahalakshmi 's solution (x,y), since if (xi,yi)=(xj,yj), with i being different from j, then xi=xj and thus di-b=dj-b and then di=dj, which is a contradiction, since di is different from dj, as a result of i being different from j.

In this way, M. Mahalakshmi 's answer gives all integer solutions of the examined equation and it gives them without any repetitions.

Once more: If ab+c=0, there's one solution and if ab+c=prime (7 in the example presented), there are two solutions for a fixed choice of sign. Those were the two cases examined. It was then found that the choice of sign of the factors of ab+c doubled the number of solutions.

If the prime decomposition of ab+c has 2 distinct factors, there are, also, two solutions.

It's when ab+c has more than 2 distinct factors that the bookkeeping becomes more complicated; as the example with 3 distinct factors shows, there are 6 solutions.

The reason is that it's necessary to establish the correspondence between the n distinct factors on the one hand and the two factors, x+b and y+a, on the other. So it's necessary to study the ways of splitting a product of n numbers into two groups.

Already for 3 distinct factors, it's not true that the solution is a linear function of the factors; the solution necessarily is quadratic in the factors, as the example shows.

That's why it's useful not to stop at the simplest cases, but look further.

What the method by M. Mahalakshmi shows is that, while the equation seems to depend on three integers, a, b and, c, in fact the parameter that matters is the combination thereof ab+c. So, if ab+c=prime, there are four solutions: two for each choice of sign. What remains to be understood is what happens, when ab+c isn't a prime, but has more than two distinct factors. Experiment shows that, for 3 factors, there are 6 solutions, of fixed sign, but what this doesn't show is, whether this means that we should write 6 as 3 x 2 or as 3!.

One way to handle the sign ambiguity is by assigning it not to the prime factorization of ab+c, but to the two factors, x+b and y+a and keep the prime factors of ab+c of fixed sign.

Stam Nicolis , irony seems to be second nature to you and this is not acceptable; you should learn to be polite.

At first, you made a statement that there exist missing solutions in the first answer, without giving any proof of it. Then you deleted that statement and you claimed that the solution given in the first answer cannot count correctly all solutions, without providing any proof of that either. When you claim something in mathematics, you have to prove it.

Once more: The solution given by M. Mahalakshmi finds the solutions only for ab+c=0 and ab+c=prime. it doesn't address the case when ab+c isn't a prime.

Furthermore, it doesn't claim to provide a method for counting the solutions. What it does do is show that the number of integer solutions of the equation doesn't depend on the three integers, a, b and c, but only on one integer, their combination ab+c. So it shows that the problem, that remains, is that of counting the-integer-solutions to the equation (x+a)(y+b)=ab+c, when ab+c is NOT a prime.

When ab+c has two distinct, coprime, factors, the number of solutions is the same as when it is a prime-so the generalization of M. Mahalakshmi solution is that, when ab+c=p, prime, then there are 2p solutions and it's possible to find them by the same approach used for the example p=7.

When ab+c has three distinct, coprime, factors, the number of solutions for a fixed sign of the factors of ab+c is 6; however it isn't true, just from that, that the total number of solutions, whatever the sign, is 12. That has to be worked out separately. And for more than three distinct, coprime, factors, how to partition the coprime factors in two groups has to be spelled out, along with the signs. It's useful to focus on the technical, not the social, issues, namely how to describe the case of n distinct, coprime, factors.

You're welcome Shouxing Qu

Stam Nicolis May I know, how do you say that the procedure given by me finds the solutions only for the cases ab+c=0 and ab+c is prime.

Because that’s what you did. Examples are useful, but what can be abstracted from them is necessary for generalization to any case.

Once more: if ab+c has two distinct coprime factors, then the equation has four solutions. This is the same as for the case when ab+c is prime.

So the method that works for the case when ab+c=prime, holds, also, when ab+c = d1 d2, with (d1, d2)=1.

When ab+c has three distinct coprime factors, then there are 6 solutions, when the three factors have all the same sign; it remains to work out what happens if they don't. And it's possible to see that there are several cases:

Since there is an odd number of factors, there are two relative signs.

If the two signs belong to the two factors, that appear together, the equations don't change, so no new solutions are found. New solutions are found when one sign belongs to the single factor and the other to the product of the two others.

So:

x+a = -d1, y+b = -d2d3 and x+a = -d2d3, y+b = -d1

and three other lines of two equations, so the total number of solutions is, indeed, 12=2x6, where 6 is the number of solutions for a fixed sign.

(But there aren't any new solutions found, when d2-> -d2 and d3->-d3, for instance. In that case, incidentally, d1-> -d1 is forbidden.)

Once more, what isn't obvious at this point is, whether the appropriate way to write 6 is as 3! or 2 x 3; is the number of solutions, for fixed sign, when there are n distinct coprime factors, 2 x n or n!? When n=3, they give the same result. For n=2, 4 = 2 x n; but that's not enough, of course, to tip the balance in favour of #solutions of fixed sign = 2 x n...

If there are four distinct coprime factors, then there are, also, solutions where

x+a = d1d2, y+b = d3d4 and x+a = d3d4, y+b = d1d2

(along with other equations)

and changing the sign of the individual factors doesn't, always, lead to new equations and, thus, new solutions. So a system of taking this into account is needed. It would be better to think about that, in order to completely solve the problem.

Dear Shouxing Qu , from a quick search, I have just found that your question was also asked, about ten years ago, on Mathematics Stack Exchange, which is, in general, a credible platform.

Here is the link: https://math.stackexchange.com/questions/449845/diophantine-equation-xyaxbyc-0

As you can see, the answers given there are the same as the two answers given here and the answer that qualified is the answer given by M. Mahalakshmi here.

It might be tempting, therefore, to conjecture that the number of solutions (counting the cases of signs) is 4n, where n is the number of distinct coprime factors of ab+c. This does seem to hold for n=1 and n=3 in fact, but there's an exception for n=2...So it's useful to understand what's going on.

The replies on mathexchange, (a) didn't solve the problem and (b) haven't corrected a mistake in the calculation.

x2-y2=N=4(ab+c)

(where the x and y here are combinations of the x and y in the original equation.)

is the Pell equation (x2-Dy2=1) for D=1, but can't be solved by the same methods as for D non-square-free, since D=1 is a perfect square.

The problem, once more, is to actually count the solutions completely. This has been done, here, only for three cases: ab+c has 1, 2 and 3 distinct coprime factors (gving 4,4 and 12 solutions respectively) and these are too few for a pattern to become clear, much less for a proof of any conjecture.

To count the solutions completely, in general, means counting the integer points on the hyperbola x2-y2=N, where N=4(ab+c), as a function of N. So there are two cases to consider, since 4 is a perfect square: (i) ab+c is a perfect square or (ii) ab+c is not a perfect square. In case (i) the problem reduces to the case of counting the integer points on the hyperbola (x/R)2-(y/R)2=1, where N=R2. This is achieved by noticing that the latter hyperbola has two integer points,

(x,y)=R(±1,0), defining the family of straight lines y = |t|x±|t|R, finding the slope, t, where the line(s) intersect(s) the hyperbola the second time and selecting and counting those values of t that give integer values for (x,y). There's still work to be done to finish the job, of course.

In case (b) things are more complicated and require more work.

The first answer on mathexchange solves the problem, actually.

Moreover, the equation xy+ax+by=c is a special type of a class of equations called quadratic Diophantine equations.

Algorithms to solve such equations may be found here http://www.numbertheory.org/php/main_pell.html .

This is a reliable page, as it is seen by the profile of the person who maintains it.

Also, a general study of these equations can be found here https://www.alpertron.com.ar/METHODS.HTM , where, again, the general solution of xy+ax+by=c is the same as that given in the first answer here.

Spiros Konstantogiannis Thank you very much for your responses, especially for your recommendation to the link of Mathematics Stack Exchange. It is a great help to me.

A solution to the problem means

#solutions of the equation = known function of the variable ab+c. Since a,b, c are integers this means a known function of the number of its prime factors.

Where's the function f(ab+c)? The ``first answer" in the mathexchange thread doesn't mention it at all-just states that it exists-that's not a solution, the whole point of the exercise is to find a way for counting the solutions!-and the ``second answer'' hasn't corrected the error in the calculation, which shows that neither has understood the problem at all.

For ab+c=prime, the answer, given by @Mahalakshmi Marikani, implies that

f(ab+c=p)=4 and this is the same when ab+c=d1d2, with (d1,d2)=1.

For ab+c =d1d2d3, explicit calculation shows that f(ab+c)=12.

The expression for f(ab+c), when ab+c = d1d2d3...dn where the factors are coprime, remains to be found. A case apart would be when ab+c=pn, with p=prime, since then the n factors are not coprime.

Apparently there are many more people that are interested in counting the integer points on regions of the plane that are bounded by hyperbolas, than are interested in counting the integer points on the hyperbolas themselves. That's sociology...

Recall that the original equation takes the form

(2u+a+b)2-(2v-(a-b))2=4(c+ab)

where u=(x+y)/2 and v=(x-y)/2.

Now assume that 4(c+ab)=R2, with R integer.

The equation takes the form

(U/R)2-(V/R)2 = 1

Now the relevant statement here is that this equation possesses only two integer solutions, namely (±1,0).

Therefore it would seem that the only integer points of the original equation are solutions of the equations

x+y+a+b = ±R and x-y-(a-b)=0

which would imply that, in this case, there exist-at most-2 (integer) solutions, whatever the value of R and of a,b,c, for which ab+c is a perfect square.

Thus f(R2/4)

However, the previous conclusion is too hasty. The reason is that rational solutions of (U/R)^2-(V/R)^2=1 are admissible, because the constraint is that x and y must be integers, not U/R and V/R!

The relevant statement here is that the rational solutions of (U/R)^2-(V/R)^2=1 are given by the expressions

U/R = (t^2+1)/(t^2-1) and V/R=2t/(t^2-1)

as mentioned in a previous entry, for t rational=p/q, with (p,q)=1. Therefore U and V will be integers, if |p^2-q^2| divides R (this is a necessary condition, not a sufficient one).

This, still, implies that there exists a finite number of solutions, just that finding it, even in this case, isn't as trivial as it might seem. So f(R2/4) remains to be found and can be greater than 2.