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Diophantine Equations Problem Set (Send corrections to [email protected]) 1. Find all integer solutions to 4x − 10y = 2. 2. Find all positive integer solutions to x2 − y 2 = 15. 3. If p, q, r are all primes satisfying pqr = 7(p + q + r), what is the value of p + q + r ? 4. Find the unique pair of positive integers that satisfies x2 + 84x + 2008 = y 2 . 5. Find all integer solutions to 4y − 615 = x2 . 6. Find all integer solutions to x2 + y = x + y 2 + 36. 7. Find all positive integer solutions to x1 + y1 = 17 . What about if you change the 7 to an arbitrary prime p? What about changing 7 to an arbitrary integer? 8. Find all pairs of integers that satisfies y = 2x2 + 5xy + 3y 2 . 9. Find all pairs of integers that satisfies 2(x2 + y 2 ) + x + y = 5xy. 10. Determine all real solutions k such that x2 − 4x − k =

√

8x2 − 32x − 8k

has precisely two distinct real solutions for x. Solution: 1. First dividing by 2, we see that we are trying to find positive integer solutions to 2x − 5y = 1. As was done in the video, we can first show that y must be odd. For if y was even, then we see that 2x − 5y is even. However this is equal to 1, an odd number which cannot happen. Thus, y is odd. Let’s write y = 2k + 1 for some integer k . Then, substituting into the equation above gives 2x − 5(2k + 1) = 1. Simplifying yields 2x = 10k + 6 and so x = 5k + 3. Thus, all integer solutions are given by (5k + 3, 2k + 1) for any integer k . 2. Factoring the left gives (x − y)(x + y) = 15. This gives x + y = 15 and x − y = 1 or x + y = 5 and x − y = 3 (since x + y is positive, negative answers are inadmissible). Adding the first two gives 2x = 16 so x = 8 and then substituting in x + y = 15 gives y = 7. Doing likewise with the second pair of equations gives 2x = 8 and so x = 4. Plugging into x + y = 5 gives y = 1. These give (x, y) = (8, 7) or (x, y ) = (4, 1). A quick check shows these satisfy the original equation. 3. (PUMAC 2013) Without loss of generality, one of p, q or r is 7 so let’s suppose p = 7 and so qr = 7 + q + r. Isolating for 7 gives qr − q − r = 7. Add 1 to both sides and factor gives q(r − 1) − (r − 1) = 8 and further (q − 1)(r − 1) = 8. This gives (q, r) as one of (2, 9), (3, 5), (5, 3), (9, 2). Only q = 3 and r = 5 works (or flip q and r). Either way, p + q + r = 7 + 5 + 3 = 15. 1

4. (AIME 2008) Completing the square gives x2 + 84x + 2008 = (x + 42)2 + 244 This is y 2 and rearranging gives 244 = y 2 − (x + 42)2 = (y − x − 42)(y + x + 42) The difference between the numbers is 82 and since x and y are positive the second term is bigger. This leaves x + y + 42 as one of 244 or 122 (the only factors of 244 larger than 82). The first case gives y − x − 42 = 1 which is a contradiction and so x +y +42 = 122 and y − x −42 = 2. Summing gives 2y = 124 or y = 62 and substituting back into x + y + 42 = 122 gives x = 18. Thus, (x, y ) = (18, 62) is the only positive solution. 5. (Math League HS 2005-2006) We can assume x > 0 since every negative solution corresponds to a positive solution. Isolating gives (2y − x)(2y + x) = 22y − x2 = 615 Hence 2y + x is a factor of 615 and is bigger than the square root of 615. These factors are 41, 123, 205, 615 and the corresponding values for 2y − x are 15, 5, 3, 1 respectively. Only 2y + x = 123 and 2y − x = 5 gives a solution of 2y+1 = 128 and hence y = 6 with x = 59 is the only solution in positive x. Thus (x, y) = (±59, 6) gives all solutions. 6. (Math League HS 2010-2011) Isolating and grouping with x2 − y 2 gives (x − y)(x + y − 1) = 36 Hence x − y is one of ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36. This means x + y − 1 is one of ±36, ±18, ±12, ±9, ±6, ±4, ±3, ±2, ±1 respectively. The sum of the two values is 2x − 1 which is odd and this leaves 2x − 1 as one of ±37, ±15, ±13. Thus, x = −18, 19, −7, 8, −6, 7. The corresponding ordered pairs (x, y) are one of (−18, −17), (−18, 18), (−7, −4),(−7, 5), (−6, −2), (−6, 3), (7, −2), (7, 3), (8, −4), (8, 5), (19, −17), (19, 18)

7. (Project Euler 108) This is equivalent to 7x + 7y = xy which is (x − 7)(y − 7) = 49. Thus, x − 7 is one of 1, 7 or 49 and correspondingly, y − 7 is one of 49, 7 or 1. Thus, the solutions for positive x and y are (x, y) as one of (8, 56), (14, 14), (56, 8). You can do something similar if you change 7 to some value for p. For n, you get lots of solution! The idea is to note that if 1/x + 1/y = 1/n, we must have that x = n + a and y = n + b for some positive a and b. Then, we have that 1 1 1 + = n+a n+b n 1 n+b+n+a = (n + a)(n + b) n n(2n + a + b) = n2 + an + bn + ab 2n2 + an + bn = n2 + an + bn + ab n2 = ab and so every positive factor of n corresponds to a solution! 2

8. (I can’t remember where I first saw this...) Rearranging by adding 0 in line 3 and grouping in line 5 gives y = 2(x2 + 2xy + y 2 ) + xy + y 2 0 = 2(x + y)2 + y (x + y − 1)

0 = 2(x + y − 1 + 1)2 + y(x + y − 1) 0 = 2(x + y − 1)2 + 4(x + y − 1) + 2 + y(x + y − 1) −2 = (x + y − 1)(2(x + y − 1) + 4 + y ) 2 = (1 − x − y)(2x + 3y + 2). Now, we are looking for integer solutions so it suffices to solve the four equations (where below we keep the signs the same in the pairs of equations) ( ( 1−x−y = ±2 1−x−y = ±1 2 + 2x + 3y = ±2 2 + 2x + 3y = ±1 Multiplying the first equation in each case by 2 and adding the pair of equations together gives the equations 4 + y = ±5

4 + y = ±4

which gives the values y as one of 1,-9, 0, -8. Plugging these into the corresponding equation (keeping the sign dependency) gives the (x, y) pairs (−2, 1), (12, −9), (0, 0), (10, −8). 9. (Awesome Math Test A) Isolating for 0 and cleverly factoring gives (2x − y − 1)(x − 2y + 1) = −1 Hence 2x − y − 1 = −1 and x − 2y + 1 = 1, or 2x − y − 1 = 1 and x − 2y + 1 = −1. In the first case, we get 2x − y = 0 and x − 2y = 0 giving x = y. In the second case, we get that 2x − y = 2 and x − 2y = −2 which also gives x = y. Back into the original, we see that 4x2 + 2x = 5x2 which gives x = 0 or x = 2 and so the solutions are (0, 0) and (2, 2). 10. (2004 COMC) Factoring and squaring both sides gives (x2 − 4x − k )2 = 8(x2 − 4x − k). Isolating for 0 gives (x2 − 4x − k)(x2 − 4x − 8 − k) = 0 In order for this to have four solutions, either the first equation has two roots and the second equation has 0, vice versa or they both have exactly one solution. The number of solutions depends on the discriminants of the quadratics. Let ∆1 and ∆2 be the discriminant of the first and second discriminants respectively. Evaluating these gives ∆1 = 16 + 4k 3

∆2 = 48 + 4k

Now, we proceed in cases Case 1: ∆1 = 0 and ∆2 = 0. This occurs when k = −4 and when k = −12, which is impossible. Case 2: ∆1 < 0 and ∆2 > 0. This occurs when k < −4 and when k > −12, which is valid. Case 3: ∆1 > 0 and ∆2 < 0. This occurs when k > −4 and when k < −12, which is impossible. Therefore, the solution is −12 < k < −4. As a sanity check, we ensure that our surd is defined on this region, that is, we need to make sure that when −12 < k < −4, we have that x2 − 4x − k ≥ 0. Completing the squares gives x2 − 4x − k = (x − 2)2 − 4 − k ≥ −4 − k and since k < −4, we have that −4 − k > 0 which confirms that values in this range indeed are in the domain of the original function and hence we do get two distinct real roots to the original equation.

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