Math Theory of Number Homework 1

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1 Math 4050 Theory of Number Homework 1 Due Wednesday, , in class Do 5 of the following 7 problems. Please only attempt 5 because I will only grade Find all rational numbers and y satisfying the equation + y = 5. [Hint: Use the change of variables u = y and v = + y and find an equation relating u and v.] Proof. Via the change of variables we get u + v = 5 + 5y = 5 so (u/5) + (v/5) = 1. We already know that the rational solutions to this equation are of the form u/5 = t/() and v/5 = (t 1)/() where t Q or t =. Thus Solving the system of equations we get y = u = 10t + y = v = 5(t 1) = (t + t 1) y = t 4t 1. Find all rational numbers and y satisfying the equation + y + 3y =. [Hint: Use the change of variables u = + y and v = y and find an equation relating u and v. Then mimick how we found all Pythagorean triples.] Proof. Via the change of variables we get u + v = + y + 3y = which has (0, 1) as a solution. If (u, v) (0, 1) is another solution let t be the -coordinate of the intersection between the -ais and the line through the pole (0, 1) and the point (u, v). Eactly as in the case of Pythagorean triples (the equation + y = 1) we get, using similar triangles, that Substituting we get v = 1 u t = u + v = u + (1 u t ) = u (1 + /t ) 4u/t + u (1 + /t ) = 4u/t 1

2 Either u = 0 or we can divide by u to get u = 4t The case u = 0 is obtained by t = 0 or t = it s incorporated in the above formula anyway. Then we get v = 1 u/t = t and these are all the rational solutions. Now we solve to get which yield all rational solutions as t Q { }. 3. Consider the diophantine equation + y = 4t y = t = t + 4t + y = t 3 + 5y + 7z = (a) Find a solution with, y, z Z. [Hint: Use the Euclidean algorithm from class.] (b) Show that if 3X + 5Y + 7Z = 0 for some integers X, Y, Z then 3 must divide Z Y. (c) Find all integral solutions to the equation. Proof. (a) From class we know that ( 1) = 1 (, 1, 0) is a solution. Or you could have used the Euclidean algorithm from the version of Bezout s formula for 3 integers. (b) Working modulo 3 we have 3 + 5y + 7z y + ( 1) z y z (mod 3) so any integral solution to 3 + 5y + 7z = 0 would have 3 y z. (c) We want to parametrize integral solutions (, y, z) to 3 + 5y + 7z =. As in class (where we did 3 + 5y = 1) we subtract from this the guessed solution to obtain the equation 3( ) + 5(y + 1) + 7z = 0 and from part (b) we know that y + 1 z (mod 3). This implies that there must eist an integer k such that y + 1 = z + 3k. Plugging back into the equation we get 0 = 3( ) + 5(y + 1) + 7z = 3( ) + 5(z + 3k) + 7z = 3( ) + 1z + 15k and dividing by 3 we get + 4z + 5k = 0 so = 4z 5k. Thus all integral solutions are of the form as k, z Z. (, y, z) = ( 4z 5k, z + 3k 1, z)

3 4. Consider the diophantine equation y = zt with, y, z, t Z. Show that there eist integers a, b, c, d such that = ab, y = cd, z = ac, t = bd. [Hint: Factor, y, z, t into primes.] Proof. First solution: If z = 0 then = 0 or y = 0. Reordering we may assume that = 0. Then take a = 0, c = y, b = t and d = 1. Similarly if y = 0 we get the desired epression. If y, z 0 then we may divide to get z = t y = q with rational q. Writing q = b/c in lowest terms (with b and c coprime) we know that z = b c t y = b c and there must eist integers a and d such that = ab, z = ac, t = db, y = dc as the numerator and denominator of a fraction are the same multiple of the numerator and denominator written in lowest terms. Second solution: First, let s assume that, y, z, t are all powers of a fied prime p. So = p X, y = p Y, z = p Z and t = p T. The equation is then p X+Y = p Z+T, i.e., X + Y = Z + T. We seek to write in other words we seek to solve = p X = ab = p A+B y = p Y = cd = p C+D z = p Z = ac = p A+C t = p T = bd = p B+D A + B = X C + D = Y A + C = Z B + D = T in the nonnegative integers. Reordering we may assume that X Z in which case the equation X + Y = Z + T implies Y T. Take B = 0. Then immediately A = X and D = T C = Y T. All of these are nonnegative solutions as desired. Now for the general case. For an integer n and a prime p write n p for the power of p that shows up in the factorization of n into primes. As prime factorization is unique if y = zt we deduce that p y p = z p t p the first case above implies that p = a p b p, y p = c p d p, z = a p c p and t = b p d p. Then take a = a p, b = b p, c = c p, and d = d p to get the desired epression. Third solution: Take a = (, z) and d = (y, t). Then = ab and z = ac for coprime integers b and c. We get y = aby = zt = act so by = ct. As b and c are coprime we deduce that b t and c y. Writing y = cd for an integer d we immediately get t = bd. 5. Show that all the solutions to the diophantine equation + y = z + t 3

4 are of the form mn + pq = mp + nq z = mp nq y = mn pq t = for integers m, n, p, q such that the above formulae yield integers. [Hint: Use the previous eercise.] Proof. Rewrite the equation as t = z y which is equivalent to ( + t)( t) = (y + z)(z y) From the previous eercise there eist integers m, n, p, q such that Solving the system yields the desired epressions. 6. In this eercise you will solve the equation with, y, z Q. + t = mn t = pq y + z = mp z y = nq + y + z = 1 (a) Suppose (, y, z) (0, 0, 1) is a solution. Let (a, b) be the point of intersection of the (y)-plane with the line through (, y, z) and (0, 0, 1). Show that a = y b = 1 z (b) Show, mimicking the procedure from the Pythagorean triples case, that every rational solution of the diophantine equation (other than (0, 0, 1)) is of the form = for rationals a, b. a b 1 + a + b y = 1 + a + b z = a + b a + b Proof. (a) Projecting to the (z)-plane, i.e., with y = 0, we get similar right triangle with legs 1 z, and 1, a. Thus 1 z = /a. Similarly we get the other equation. (b) Note that /a = y/b y = b/a. We have Either = 0 or = and 1 = + y + z = + b /a + (1 /a) (1 + b /a + 1/a ) = /a a a +b +1 and the former case is a special eample of the latter. Now compute y = b/a = b a + b + 1 z = 1 /a = a + b 1 a + b + 1 4

5 7. Suppose two of the integers a 1, a,..., a n are coprime. Suppose 1 = u 1,..., n = u n is an integral solution to the diophantine equation Find all the other solutions. [Hint: Cf. eercise 3.] a a n n = b Proof. As in Eercise 3 we can rewrite the equation as ai i = d = a i u i or equivalently ai ( i u i ) = 0 Suppose for simplicity that a 1 and a are coprime (otherwise simply reorder the indices). Then a is invertible modulo a 1 and there eists b Z such that a b 1 (mod a 1 ). If 1,..., n is a solution then reducing modulo a 1 we get and multiplying with b we get Thus we may write a i ( i u i ) 0 (mod a 1 ) i= u = ba i ( i u i ) (mod a 1 ) u = a 1 k ba i ( i u i ) for some integer k. Plugging it back into the equation we get a 1 ( 1 u 1 ) + a (a 1 k ba i ( i u i )) + 1 u 1 = a k + a i ( i u i ) = 0 ba 1 a 1 a i ( i u i ) where all coefficients are now integers as ba 1 (mod a 1 ). Thus every solution is of the form ( 1,..., n ) where ba 1 1 = u 1 a k + a i ( i u i ) a 1 for k, 3,..., n Z. = u + a 1 k ba i ( i u i ) 5