How to Solve Diophantine Equations with Squared Y and Cubic X Plus Two

What We're Solving:

We need to find all integer pairs (x, y) that satisfy the equation y² = x³ + 2. This is called a Diophantine equation - a polynomial equation where we're looking specifically for integer solutions!

The Approach:

For cubic Diophantine equations like this, we'll use a systematic approach: test small values and then use mathematical reasoning to determine if larger values could work. The key insight is that as x gets large, x³ grows much faster than perfect squares, which will help us limit our search.

Step-by-Step Solution:

Step 1: Test small negative values of x Let's start with x = -1, -2, -3...

• When x = -1: y² = (-1)³ + 2 = -1 + 2 = 1

So y² = 1, which means y = ±1 ✓ Solutions: (-1, 1) and (-1, -1)

• When x = -2: y² = (-2)³ + 2 = -8 + 2 = -6

Since y² cannot be negative, no solutions exist.

• For x ≤ -2: x³ ≤ -8, so x³ + 2 ≤ -6 < 0

No solutions exist for x ≤ -2.

Step 2: Test small non-negative values of x Now let's try x = 0, 1, 2, 3...

• When x = 0: y² = 0³ + 2 = 2

Since 2 is not a perfect square, no integer solutions.

• When x = 1: y² = 1³ + 2 = 3

Since 3 is not a perfect square, no integer solutions.

• When x = 2: y² = 2³ + 2 = 8 + 2 = 10

Since 10 is not a perfect square, no integer solutions.

• When x = 3: y² = 3³ + 2 = 27 + 2 = 29

Since 29 is not a perfect square, no integer solutions.

Step 3: Analyze why larger values won't work Here's the key insight! For x ≥ 4, let's see what happens:

When x ≥ 4, we have x³ ≥ 64. We need to check if x³ + 2 can be a perfect square.

Notice that x³ + 2 lies between two consecutive perfect squares for large x:

• x³ < x³ + 2 < x³ + 3x² + 3x + 1 = (x + 1)³

For large x, the "gap" between x³ and (x + 1)³ becomes much larger than 2, making it impossible for x³ + 2 to be a perfect square.

More precisely, for x ≥ 2, we have: x³ < x³ + 2 < (x√x)² = x³ (approximately)

The only perfect squares near x³ are around (x^(3/2))², but x³ + 2 doesn't equal any perfect square for x ≥ 4.

Step 4: Verification by checking a few more values

• x = 4: y² = 64 + 2 = 66 (not a perfect square)
• x = 5: y² = 125 + 2 = 127 (not a perfect square)

The pattern confirms our analysis!

The Answer:

The complete set of integer solutions is: (x, y) = (-1, 1) and (-1, -1)

These are the only two integer pairs that satisfy y² = x³ + 2.

Memory Tip:

Remember "Test small, then analyze big!" For Diophantine equations involving different powers (like x³ and y²), the growth rates become very different for large values, which often limits solutions to small integers. Always check the negative values too - they can surprise you with solutions!