Problem Set 1 Solution Proof. We use case analysis. Since m 23, one of four cases must hold 1. m=3. There are now four subcase (a)n=3. Rewriting the equation in the form and subtituting in m= n= 3 implies that e= 6, which is the first solution (b)n=4. Substituting the values of m and n into the equation show that e= 12, which is the second solution (c)n=5. Substituting into the equation shows that e 30, which is the third solution (d)n≥6. This implies: 11 Thus, the left side of the equation is strictly less than the right for all e >0,so there are no solutions in this case 2. m= 4. There are two subcases (a)n=3. Subsituting gives e= 12, which is the fourth solution (b)n≥4. This implie 1+1<1 Again, the left side of the equation is strictly less than the right for alle>0,so there are no solutions in this case 3. m=5. There are two subcases (a)n=3. Subsituting gives e= 30, which is the fifth solution (b)n≥4. This implies: 11111 Again, the equation can not hold, so there are no solutions in this cas 4.m≥6. This implies: 1 1111 m+n=6+3=2 Once more the equation can not hold, so there are no solutions in this case.Problem Set 1 5 Solution. Proof. We use case analysis. Since m ≥ 3, one of four cases must hold: 1. m = 3. There are now four subcases: (a) n = 3. Rewriting the equation in the form 1 e = 1 + 1 nm − 1 2 and subtituting in m = n = 3 implies that e = 6, which is the first solution. (b) n = 4. Substituting the values of m and n into the equation show that e = 12, which is the second solution. (c) n = 5. Substituting into the equation shows that e = 30, which is the third solution. (d) n ≥ 6. This implies: 1 1 1 1 1 m + n ≤ 3 + 6 = 2 Thus, the left side of the equation is strictly less than the right for all e > 0, so there are no solutions in this case. 2. m = 4. There are two subcases: (a) n = 3. Subsituting gives e = 12, which is the fourth solution. (b) n ≥ 4. This implies: 1 1 1 1 1 + = m + n ≤ 4 4 2 Again, the left side of the equation is strictly less than the right for all e > 0, so there are no solutions in this case. 3. m = 5. There are two subcases: (a) n = 3. Subsituting gives e = 30, which is the fifth solution. (b) n ≥ 4. This implies: 1 1 1 1 1 + < m + n ≤ 5 4 2 Again, the equation can not hold, so there are no solutions in this case. 4. m ≥ 6. This implies: 1 1 1 1 1 + = m + n ≤ 6 3 2 Once more, the equation can not hold, so there are no solutions in this case