1. The problem statement, all variables and given/known data Give a proof by cases to show that the equation 5x^2 + 4y^3 = 51 does not have solutions x, y ∈ Z+ 3. The attempt at a solution My wording is very awkward, i am hoping that i can get some advice on it. To prove there are no solutions x, y ∈ Z+ we first divide into cases. Note that 5x² cannot be equal to 4y³ as 51 is an odd number and an odd number divided by 2 does not give a result ∈ Z+. Case 1: 5x² < 4y³. In this case, only values of y that result in a value from 4y³ larger than 51 / 2 are usable. It follows that only the possible value for y in this case is 2. Therefore there must be a value for 5x² which results in 19. Testing with x=1 and x=2 we see that there is no value for 5x² which results in 19. Thus we conclude there are no solutions in this case. Case 2: 5x² > 4y³. In this case, only values of x that result in a value from 5x² larger than 51 / 2 are usable. It follows that the only possible value for x in this case is 3. Therefore we need a value for 4y³ which results in 6. Testing with y=1 and y=2 we see that there is clearly no value for 4y³ which results in 6. There are no solutions in this case. Thus we conclude that since we could not find a solution in either case, there is no solution.