To find the integer solutions for the equation $ x^3 + y^5 = 40 $, where $ x $ and $ y $ are integers, we can start by testing various values for $ y $ and calculating the corresponding $ x $. 1. **For $ y = 0 $**: \[ x^3 + 0^5 = 40 \implies x^3 = 40 \implies x \approx 3.4 \quad \text{(no integer solution)} \] 2. **For $ y = 1 $**: \[ x^3 + 1^5 = 40 \implies x^3 + 1 = 40 \implies x^3 = 39 \implies x \approx 3.4 \quad \text{(no integer solution)} \] 3. **For $ y = 2 $**: \[ x^3 + 2^5 = 40 \implies x^3 + 32 = 40 \implies x^3 = 8 \implies x = 2 \quad \text{(solution: } (2, 2) \text{)} \] 4. **For $ y = 3 $**: \[ x^3 + 3^5 = 40 \implies x^3 + 243 = 40 \implies x^3 = -203 \quad \text{(no integer solution)} \] 5. **For $ y = -1 $**: \[ x^3 + (-1)^5 = 40 \implies x^3 - 1 = 40 \implies x^3 = 41 \implies x \approx 3.4 \quad \text{(no integer solution)} \] 6. **For $ y = -2 $**: \[ x^3 + (-2)^5 = 40 \implies x^3 - 32 = 40 \implies x^3 = 72 \implies x \approx 4.16 \quad \text{(no integer solution)} \] Continuing similarly for negative values of $ y $ and larger positive values will yield no further integer solutions. After checking the values sequentially, we conclude that the only integer solution to the equation $ x^3 + y^5 = 40 $ is: \[ \boxed{(2, 2)} \]

To find integer solutions for the equation $ x^2 + y^3 = 44 $, we can systematically check various integer values for $ y $ and solve for $ x $. 1. **For $ y = 0 $**: \[ x^2 + 0^3 = 44 \implies x^2 = 44 \implies x \approx 6.63 \quad \text{(no integer solution)} \] 2. **For $ y = 1 $**: \[ x^2 + 1^3 = 44 \implies x^2 + 1 = 44 \implies x^2 = 43 \implies x \approx 6.56 \quad \text{(no integer solution)} \] 3. **For $ y = 2 $**: \[ x^2 + 2^3 = 44 \implies x^2 + 8 = 44 \implies x^2 = 36 \implies x = 6 \quad \text{(solution: } (6, 2) \text{)} \] 4. **For $ y = 3 $**: \[ x^2 + 3^3 = 44 \implies x^2 + 27 = 44 \implies x^2 = 17 \implies x \approx 4.12 \quad \text{(no integer solution)} \] 5. **For $ y = 4 $**: \[ x^2 + 4^3 = 44 \implies x^2 + 64 = 44 \implies x^2 = -20 \quad \text{(no integer solution)} \] Next, we will check negative values for $ y $: 6. **For $ y = -1 $**: \[ x^2 + (-1)^3 = 44 \implies x^2 - 1 = 44 \implies x^2 = 45 \implies x \approx 6.71 \quad \text{(no integer solution)} \] 7. **For $ y = -2 $**: \[ x^2 + (-2)^3 = 44 \implies x^2 - 8 = 44 \implies x^2 = 52 \implies x \approx 7.21 \quad \text{(no integer solution)} \] 8. **For $ y = -3 $**: \[ x^2 + (-3)^3 = 44 \implies x^2 - 27 = 44 \implies x^2 = 71 \implies x \approx 8.43 \quad \text{(no integer solution)} \] 9. **For $ y = -4 $**: \[ x^2 + (-4)^3 = 44 \implies x^2 - 64 = 44 \implies x^2 = 108 \implies x \approx 10.39 \quad \text{(no integer solution)} \] After checking integer values for $ y $ from $-4$ to $4$, the only integer solution to the equation $ x^2 + y^3 = 44 $ is: \[ \boxed{(6, 2)} \]

To find integer solutions for the equation $ x^3 + y^2 = 17 $, we can test various integer values for $ y $ and compute the corresponding $ x $. 1. **For $ y = 0 $**: \[ x^3 + 0^2 = 17 \implies x^3 = 17 \implies x \approx 2.57 \quad \text{(no integer solution)} \] 2. **For $ y = 1 $**: \[ x^3 + 1^2 = 17 \implies x^3 + 1 = 17 \implies x^3 = 16 \implies x = 2 \quad \text{(solution: } (2, 1) \text{)} \] 3. **For $ y = 2 $**: \[ x^3 + 2^2 = 17 \implies x^3 + 4 = 17 \implies x^3 = 13 \implies x \approx 2.35 \quad \text{(no integer solution)} \] 4. **For $ y = 3 $**: \[ x^3 + 3^2 = 17 \implies x^3 + 9 = 17 \implies x^3 = 8 \implies x = 2 \quad \text{(solution: } (2, 3) \text{)} \] 5. **For $ y = 4 $**: \[ x^3 + 4^2 = 17 \implies x^3 + 16 = 17 \implies x^3 = 1 \implies x = 1 \quad \text{(solution: } (1, 4) \text{)} \] 6. **For $ y = 5 $**: \[ x^3 + 5^2 = 17 \implies x^3 + 25 = 17 \implies x^3 = -8 \implies x = -2 \quad \text{(solution: } (-2, 5) \text{)} \] 7. **For larger values of $ y $** (i.e., $ y \geq 6 $), $ y^2 $ exceeds 17, making $ x^3 $ negative, leading to no positive integer solutions. Next, we can also check negative values for $ y $: 8. **For $ y = -1 $**: \[ x^3 + (-1)^2 = 17 \implies x^3 + 1 = 17 \implies x^3 = 16 \implies x = 2 \quad \text{(solution: } (2, -1) \text{)} \] 9. **For $ y = -2 $**: \[ x^3 + (-2)^2 = 17 \implies x^3 + 4 = 17 \implies x^3 = 13 \implies x \approx 2.35 \quad \text{(no integer solution)} \] 10. **For $ y = -3 $**: \[ x^3 + (-3)^2 = 17 \implies x^3 + 9 = 17 \implies x^3 = 8 \implies x = 2 \quad \text{(solution: } (2, -3) \text{)} \] 11. **For $ y = -4 $**: \[ x^3 + (-4)^2 = 17 \implies x^3 + 16 = 17 \implies x^3 = 1 \implies x = 1 \quad \text{(solution: } (1, -4) \text{)} \] Thus, the integer solutions for the equation $ x^3 + y^2 = 17 $ are: \[ \boxed{(2, 1), (2, 3), (1, 4), (-2, 5), (2, -1), (2, -3), (1, -4)} \]

?^3 + ?^5 = 40

To find the integer solutions for the equation $x^{3} + y^{5} = 40$ , where $x$ and $y$ are integers, we can start by testing various values for $y$ and calculating the corresponding $x$ .

For $y = 0$ : $x^{3} + 0^{5} = 40 ⟹ x^{3} = 40 ⟹ x \approx 3.4 (no integer solution)$
For $y = 1$ : $x^{3} + 1^{5} = 40 ⟹ x^{3} + 1 = 40 ⟹ x^{3} = 39 ⟹ x \approx 3.4 (no integer solution)$
For $y = 2$ : $x^{3} + 2^{5} = 40 ⟹ x^{3} + 32 = 40 ⟹ x^{3} = 8 ⟹ x = 2 (solution: (2, 2))$
For $y = 3$ : $x^{3} + 3^{5} = 40 ⟹ x^{3} + 243 = 40 ⟹ x^{3} = - 203 (no integer solution)$
For $y = - 1$ : $x^{3} + (- 1)^{5} = 40 ⟹ x^{3} - 1 = 40 ⟹ x^{3} = 41 ⟹ x \approx 3.4 (no integer solution)$
For $y = - 2$ : $x^{3} + (- 2)^{5} = 40 ⟹ x^{3} - 32 = 40 ⟹ x^{3} = 72 ⟹ x \approx 4.16 (no integer solution)$

Continuing similarly for negative values of $y$ and larger positive values will yield no further integer solutions. After checking the values sequentially, we conclude that the only integer solution to the equation $x^{3} + y^{5} = 40$ is:

$(2, 2)$

?^2 + ?^3 = 44

To find integer solutions for the equation $x^{2} + y^{3} = 44$ , we can systematically check various integer values for $y$ and solve for $x$ .

For $y = 0$ : $x^{2} + 0^{3} = 44 ⟹ x^{2} = 44 ⟹ x \approx 6.63 (no integer solution)$
For $y = 1$ : $x^{2} + 1^{3} = 44 ⟹ x^{2} + 1 = 44 ⟹ x^{2} = 43 ⟹ x \approx 6.56 (no integer solution)$
For $y = 2$ : $x^{2} + 2^{3} = 44 ⟹ x^{2} + 8 = 44 ⟹ x^{2} = 36 ⟹ x = 6 (solution: (6, 2))$
For $y = 3$ : $x^{2} + 3^{3} = 44 ⟹ x^{2} + 27 = 44 ⟹ x^{2} = 17 ⟹ x \approx 4.12 (no integer solution)$
For $y = 4$ : $x^{2} + 4^{3} = 44 ⟹ x^{2} + 64 = 44 ⟹ x^{2} = - 20 (no integer solution)$

Next, we will check negative values for $y$ :

For $y = - 1$ : $x^{2} + (- 1)^{3} = 44 ⟹ x^{2} - 1 = 44 ⟹ x^{2} = 45 ⟹ x \approx 6.71 (no integer solution)$
For $y = - 2$ : $x^{2} + (- 2)^{3} = 44 ⟹ x^{2} - 8 = 44 ⟹ x^{2} = 52 ⟹ x \approx 7.21 (no integer solution)$
For $y = - 3$ : $x^{2} + (- 3)^{3} = 44 ⟹ x^{2} - 27 = 44 ⟹ x^{2} = 71 ⟹ x \approx 8.43 (no integer solution)$
For $y = - 4$ : $x^{2} + (- 4)^{3} = 44 ⟹ x^{2} - 64 = 44 ⟹ x^{2} = 108 ⟹ x \approx 10.39 (no integer solution)$

After checking integer values for $y$ from $- 4$ to $4$ , the only integer solution to the equation $x^{2} + y^{3} = 44$ is:

$(6, 2)$

?^3 + ?^2 = 17

To find integer solutions for the equation $x^{3} + y^{2} = 17$ , we can test various integer values for $y$ and compute the corresponding $x$ .

For $y = 0$ : $x^{3} + 0^{2} = 17 ⟹ x^{3} = 17 ⟹ x \approx 2.57 (no integer solution)$
For $y = 1$ : $x^{3} + 1^{2} = 17 ⟹ x^{3} + 1 = 17 ⟹ x^{3} = 16 ⟹ x = 2 (solution: (2, 1))$
For $y = 2$ : $x^{3} + 2^{2} = 17 ⟹ x^{3} + 4 = 17 ⟹ x^{3} = 13 ⟹ x \approx 2.35 (no integer solution)$
For $y = 3$ : $x^{3} + 3^{2} = 17 ⟹ x^{3} + 9 = 17 ⟹ x^{3} = 8 ⟹ x = 2 (solution: (2, 3))$
For $y = 4$ : $x^{3} + 4^{2} = 17 ⟹ x^{3} + 16 = 17 ⟹ x^{3} = 1 ⟹ x = 1 (solution: (1, 4))$
For $y = 5$ : $x^{3} + 5^{2} = 17 ⟹ x^{3} + 25 = 17 ⟹ x^{3} = - 8 ⟹ x = - 2 (solution: (- 2, 5))$
For larger values of $y$ (i.e., $y \geq 6$ ), $y^{2}$ exceeds 17, making $x^{3}$ negative, leading to no positive integer solutions.

Next, we can also check negative values for $y$ :

For $y = - 1$ : $x^{3} + (- 1)^{2} = 17 ⟹ x^{3} + 1 = 17 ⟹ x^{3} = 16 ⟹ x = 2 (solution: (2, - 1))$
For $y = - 2$ : $x^{3} + (- 2)^{2} = 17 ⟹ x^{3} + 4 = 17 ⟹ x^{3} = 13 ⟹ x \approx 2.35 (no integer solution)$
For $y = - 3$ : $x^{3} + (- 3)^{2} = 17 ⟹ x^{3} + 9 = 17 ⟹ x^{3} = 8 ⟹ x = 2 (solution: (2, - 3))$
For $y = - 4$ : $x^{3} + (- 4)^{2} = 17 ⟹ x^{3} + 16 = 17 ⟹ x^{3} = 1 ⟹ x = 1 (solution: (1, - 4))$

Thus, the integer solutions for the equation $x^{3} + y^{2} = 17$ are:

$(2, 1), (2, 3), (1, 4), (- 2, 5), (2, - 1), (2, - 3), (1, - 4)$