Math Olympiad Problem!!!?

Question

Determine a positive constant c such that the equation
xy^2 − y^2 − x + y = c
has precisely three solutions (x, y) in positive integers.

P.S. xy^2 = x*(y^2)

Answer 1

I read the question too late, but the problem caught my attention and I added a star to it.
Let's for convenience rewrite xy² - y² - x + y = c as
(x-1)y² - x + 1 + y = c + 1, now denoting
z = x-1 and d = c+1 we obtain
z(y² - 1) + y = d, or, following Dr D's approach
z = (d - y)/(y² - 1), here is of course the obvious solution y=d, z=0, so the question is for what d among y=2,3,..,sqrt(d); z=(d-2)/3,..,2,1 there are exactly 2 more solutions?
Solving it for y also we need the discriminant 1 + 4z(z+d) to be a perfect square, that leads to the requirement z(z+d) to be twice a triangular number:
z(z + d) = a(a + 1) for a = 1,2,...
Let T_a = a(a + 1) is a twice triangular /2,6,12,20,30,42, etc./, then a solution would be
d = T_a/z - z, yz = a if z is a divisor of a.

So we'll have a solution of our problem if there exist exactly 2 twice triangular numbers T_a and T_b, such that
d = T_a/z' - z' = T_b/z" - z" for some z', z" and of course
y' = a/z', y" = b/z".

Taking 12 and 42 for example we obtain
d = 12/1 - 1 = 42/3 - 3 = 11, y'=2, z'=3, y"=3, z"=1 - the solution in Dr D's answer /c = d-1 = 10, x = z+1/.

If we have a solution with the least possible y=2, then d = 3z + 2 and trying to find another with same y=2 and z' = z+2 leads to d' = 3z' + 2 = d + 6, which turns out in many cases to be successful, this way and using the approach above, I found the following solutions /not counting the obvious/:
d=11; y'=2; z'=3; y"=3; z"=1;
d=19; y'=3; z'=2; y"=4; z"=1;
d=29; y'=2; z'=9; y"=5; z"=1;
d=35; y'=2; z'=11; y"=3; z"=4;
d=41; y'=2; z'=13; y"=6; z"=1; (* * * *)
d=53; y'=2; z'=17; y"=5; z"=2;
d=59; y'=2; z'=19; y"=3; z"=7;
d=71; y'=2; z'=23; y"=8; z"=1;
d=77; y'=2; z'=25; y"=5; z"=3;
d=83; y'=2; z'=27; y"=3; z"=10;
d=89; y'=2; z'=29; y"=9; z"=1;

I suppose the solutions to be infinitely many, but neither I can prove it for the moment, nor I know the description of the set of all "d". Sorry, not much a success, I'll need more time, but it's about to expire. I kindly ask everybody who will have more about this intriguing problem, to let me know.

P.S.(EDIT) Alexey: "So c should not be factored in more than 3 different ways" - that's wrong, in Your answer "k" is simply x-1 (or "z" in my answer above) and some factorings of "c" may not generate integer "k", for example
40 = 1*40 = (2 - 1)(13*2 + 13 + 1) =
= 2*20 = 4*10 =
= 5*8 = (6 - 1)(1*6 + 1 + 1), so 2*20 and 4*10 DO NOT yield integer k, look at (* * * *) above, but c=40, being factored in more than 3 ways, is a valid value!

Answer 2

>> 1) (1 / 15) < (1 / 2) (3 / 4) (5 / 6) ... (99 / 100) < (1 / 10) If you multiply it out (a one-liner in Perl), you get 1/15 = 0.0666... < 0.079589... < 0.1 = 1/10 Symbolically, the denominator is 2^50 * 50! and the numerator is 100! / (denominator), so overall the product is P = 100! / (2^100 * 50! * 50!) = "100 Choose 50" / 2^100 So P is the probability of tossing a fair coin 100 times and getting exactly 50 heads. You can approximate it with either Sitrling's approximation for the factorial or with the normal distribution approximation to the Binomial(N = 100, p = q = 1/2) distribution. The latter would be Prob(49.5 <= X <= 50.5) when X is normally distributed with mean Np = 50 and variance Npq = 25 (and so standard deviation 5). Thus we want the probability that a normal is within +/- 0.5/5 = +/- 0.1 standard devi ations from the mean, which is about 0.079656, between 1/15 and 1/10. The former, Stirling's approximation, says 1 = lim n -> pos. inf. n! / (sqrt(2 pi n) (n/e)^n) So P is approx. sqrt(2 pi 100) (100/e)^100 / (2^100 * (sqrt(2 pi 50) (50/e)^50)^2) = 10 sqrt(2 pi) (100/e)^100 / (sqrt(2 pi) sqrt(2 pi) sqrt(50) sqrt(50) 2^100 (50/e)^100) = (10 / 50) (1 / sqrt(2 pi) (100/e)^100 / ((100/e)^100) = 1 / (5 sqrt(2 pi)) 2 pi is 6.28... is a little over 6.25 = 2.5^2, so sqrt(2 pi) is a little over 2.5, and 5 sqrt(2 pi) is a little over 12.5. So 10 < 5 sqrt(2 pi) < 15, therefore 1/15 < (1 / (5 sqrt(2 pi))) approx= P < 1/10. >> 2) Consider a 25 x 25 square grid in each of whose 625 squares is placed either 'a+1' or 'a-1'. Suppose a(i) denotes the product of all the numbers in the (i)th row and b(j) denotes the product of all the numbers in the (j)th column. Prove that a(1) + b(1) + a(2) + b(2) + ... + a(25) + b(25) is not equal to 0. I think you stated this wrong. If you use 'a+1' and 'a-1' as if a were a variable, then the specified sum is a polynomial of degree at most 25 in a, and so there will be values of a which make the sum 0 (25 complex root, not necessarily distinct; at least one real root because the coefficients are real and the degree is odd). For example, let a = 1, and assign enough a-1 labels so that each row and each column contains at least one a-1 = 0. Instead, I think they meant to say you assign one of the values { -1, 1 } to each of the 625 squares. That is, "assign a '+1' or a '-1'." Anyway, with that interpretation, every row product and every column product is 1 or -1. Start with all 1's. Each row product a(i) and each column product b(j) is 1. Flip 1's to -1's one at a time and watch what happens to the sum of the a(i) + b(j). At each flip, one of the a(i) changes from -1 to 1 or vice versa, and the sum a(1) + ... + a(25) either goes up 2 or down 2. Likewise exactly one of the b(j) changes from 1 to -1 or vice versa, causing the sum b(1) + ... + b(25) to either go up 2 or go down 2. So the overall sum changes by either of + 2 + 2 = 4 + 2 - 2 = 0 - 2 + 2 = 0 - 2 - 2 = -4 As each 1 is flipped to a minus 1, the sum mod 4 is left unchanged. Since the sum with all 1's was 50 = 2 mod 4, the sum for every configuration of 1's and -1's is 2 mod 4. It is never 0. >> 3) If p and p^2 + 2 are primes, prove that p^3 + 2 is also prime. P cannot be 2 because then p^2 + 2 = 4 + 2 = 6 is not prime. p can be 3 because then p^2 + 2 = 9 + 2 = 11 and p^3 + 2 = 27 + 2 = 29 are all primes. Suppose p is a prime greater than 3. p is not divisible by 3, so either p = 1 mod 3 or p = 2 mod 3. In either case, p^2 + 2 = 1 + 2 = 3 = 0 mod 3 and so 3 divides p^2 + 2 and so p^2 + 2 is not prime. So the only time both p and p^2 + 2 are both prime is when p = 3, in which case p^3 + 2 is also a prime. Dan

Answer 3

Clearly c must be an integer, becaues integral (x,y) will not give a non integral RHS.

We can rewrite the equation:
x = (y^2 - y + c) / (y^2 - 1)

We can observe that y=1 is an asymptote.
The numerator has no roots for c >= 1 and is always positive.

We can also observe that at x=1, y = c + 1 which is an integer.
So (1, c+1) is definitely an integral solution pair.

Playing around with some numbers:
if y=2, x = (2+c)/3
If y=3, x = (c+6)/8

The lowest value of c that gives integral x in both those cases is c=10, which gives integral solutions (1,11); (2,3); (4,2).

There are no integral pairs for y < 2, hence there are no integral pairs for x > 4. There are no integral pairs for x < 1, and we can verify that x=3 gives a non integral y.

Hence c=10 gives only 3 integral solution pairs.

Answer 4

x = (y^2 - y + c) / (y^2 - 1) = 1 + (c + 1 - y)/(y^2 - 1)
Second term is integer when c = k(y^2 - 1) + y - 1 = (y -1)(ky + k + 1)
So c should not be factored in more than 3 different ways.