If the number (a^2 + b^2)/(1+a*b) with a,b integers is a positive integer, then it is a perfect square. (The stipulation of *positive* integer is required, because we have integers a=1, b=-2 such that (a^2 + b^2)/(1+ab) equals the integer -5, but this is not a perfect square.) To prove this, first note that, for any positive integer N, if the equation x^2 + y^2 --------- = N 1 + xy is an integer, with x > y (clearly x cannot equal y except when x=y=1), then we have the following polynomial with integer coefficients x^2 - (Ny)x + (y^2 - N) = 0 The two roots of this equation x1 and x2 satisfy the relation x1 + x2 = Ny x1 x2 = y^2 - N Hence if (x1,y) is a solution, then so is (x2,y) where x2 = Ny - x1. Also, since x1 exceeds y, we know that y exceeds x2, as is clear from the right hand expression above. Repeating this process, we have a strictly decreasing sequence of integers given by s[0] = x1, s[1] = y, s[k] = N s[k-1] - s[k-2] We know that s[k]^2 + s[k+1]^2 ----------------- = N 1 + s[k]s[k+1] for k=1,2,3,... If s[k+1] = 0 for some index k, then N = s[k]^2, so N is a square. The key is to show that this sequence MUST pass through 0. To prove this, suppose the sequence does not pass through zero. It follows that, since the sequence is strictly decreasing, it must contain two consecutive integers with opposite signs. Thus the quantity (x^2 + y^2)/(1+xy) must be either infinite (if xy=-1) or negative (if xy < -1), contradicting that fact that this quantity equals N, which is a finite positive integer. This completes the proof. Notice that if the quantity N = (x^2 + y^2)/(1+xy) is a negative integer for integers x,y, then N=-5. There are exactly two distinct infinite families of solutions in this case, given by the following sequences (with alternating signs) 1 2 9 43 206 987 ... 1 3 14 67 321 1538 ... each of which satisfies the recurrence s[k] = 5s[k-1] - s[k-2]. We can generalize the proposition to cover positive integers of the form x^2 + Kxy + y^2 --------------- = N 1 + xy for all integers N greater than K, where K is a positve integer not equal to 2. (We exclude 2 for reasons that will be explained below.) In this case we have the polynomial x^2 - (Ny-Ky)x + (y^2 - N) = 0 which shows that if (x1,y) is a solution, then (x2,y) is also a solution, where x2 = (N-K)y - x1. Also, if x1 exceeds y, then x2 is less than y. If we try to prove that every integer N (greater than K) given by this equation is a square, we can proceed just as before, except that now, depending on the value of K, it is possible to have x,y with opposite signs yielding an integer value of N greater than K. We excluded the case K=2 because in that case we can achieve ANY value of N by setting y=1 and x=N-1. This is not ruled out by the "descent" argument because all these cases descend back to the pair 1,-1, which gives the indeterminate 0/0, and allows for any value of N. So, excluding that special case, let X,Y denote the absolute values of x and y, and observe that if x and y have opposite signs the expression for N can be written in the form X^2 + Y^2 - K N = K - ------------- XY-1 which shows that x^2 + y^2 must be less than K in order for N to exceed K. Thus there are just a finite number of pairs (x,y) with opposite sign that can yield a value of N greater than K. These are the only possible non-square values of N. The smallest possible pair is 1,-2 (which is equivalent to 2,-1), in which case we have N = (5-2K)/-1, which we require to be greater than K. This implies the inequality K > 5. Hence, any integer value of N (greater than K) must be a square for K = 0, 1, 3, 4, or 5. For K=6 the pair 1,-2 gives N=7, so every integer value of N greater than 6 is either a square or equal to 7. More generally, every value of K greater than 5 allows the pair 1,-2, so we always have the N value of N = 2K-5. For example, with K=7 the pair 1,-2 gives N=9. However, in this particular case the "exceptional" N value also happens to be a square, so we can say that any integer N greater than 7 is a square. This occurs for K = 7, 15, and 115, so the non-negative integers K such that every positive integer N of the form (x^2 + Kxy + y^2)/(1+xy) greater than K is a square is 0, 1, 3, 4, 5, 7, 15, 115 The number of minimal pairs x,y with opposite sign yielding integer values of N tends to increase as K increases, but it appears that there are infinitely many K for which the only reduced pair is 1,-2. The first such K values are 6, 7, 8, 9, 10, 13, 15, 19, 21, 25, 31, 37, 39, 45, 49, 61, 81, 85, 99, 109, 115, 129, 141, 145, 151, 159, 169, 175, 229, 285, 295, 319, 325, 361, 375, 445, 501, ... (This list includes the squares of 3, 5, 7, 9, 13, 19, and 63.) These numbers are the result of a progressive sieve, analogous to the prime sieve. For example, every term greater than 10 must not be divisible by 2, because otherwise it would give an integer N for (3K-10)/2 based on the pair 1,-3. Likewise from the pair 2,-2 we see that every term greater than 8 must not be congruent to 2 modulo 3, because otherwise it would give an integer N for (4K-8)/3. Here is a short table of the expressions that must not be integers for sufficiently large "prime K" values -1 -2 -3 -4 -5 1 - (2K-5)/1 (3K-10)/2 ( 4K-17)/ 3 ( 5K-26)/ 4 2 (4K-8)/3 (6K-13)/5 ( 8K-20)/ 7 (10K-29)/ 9 3 (9K-18)/8 (12K-25)/11 (15K-34)/14 4 (16K-32)/15 (20K-41)/19 5 (25K-50)/24 In each case the expression (AK-B)/(A-1) implies that for K values greater than B we must exclude those such that K = B (mod A-1). In other words, the sieve excludes every number greater than q = x^2 + y^2 congruent to q mod (xy-1).

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