Proof by Cases: No Solutions for 5x^2 + 4y^3 = 51 | Z+ x, y

[twoski]

Homework Statement

Give a proof by cases to show that the equation

5x^2 + 4y^3 = 51

does not have solutions x, y ∈ Z+

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.

 

[KataKoniK]

Hi there! Your wording is actually quite clear and concise. However, I would suggest a few changes to make it more mathematically rigorous:

To prove that there are no solutions x, y ∈ Z+ to the equation 5x^2 + 4y^3 = 51, we will use proof by cases. First, note that 5x^2 cannot be equal to 4y^3 as 51 is an odd number and an odd number divided by 2 does not result in a value ∈ Z+.

Case 1: 5x^2 < 4y^3. In this case, y must be a positive integer that results in a value larger than 51/2 for 4y^3. The only possible value for y in this case is 2, which would result in 4y^3 = 32. However, there is no value for x that would result in 5x^2 = 19. Hence, there are no solutions in this case.

Case 2: 5x^2 > 4y^3. In this case, x must be a positive integer that results in a value larger than 51/2 for 5x^2. The only possible value for x in this case is 3, which would result in 5x^2 = 45. However, there is no value for y that would result in 4y^3 = 6. Therefore, there are no solutions in this case.

Since we could not find a solution in either case, we can conclude that there are no solutions x, y ∈ Z+ to the equation 5x^2 + 4y^3 = 51.