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Find all solutions to the equation x(x + 1)(x + 7)(x + = y^2 with x,y ∈ Z.

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Find all solutions to the equation x(x + 1)(x + 7)(x + 8 ) = y^2 with x,y ∈ Z.

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To find all integer solutions (x, y) to the equation x(x+1)(x+7)(x+n)=y^2, where n is an arbitrary integer, we can start by analyzing the possible values of x.
Since the product of four consecutive integers is always divisible by 4, one of these integers must be a perfect square to ensure that the entire product is a perfect square. Let's denote this perfect square as k^2, where k is an integer.

So, we have x(x+1)(x+7)(x+n)=k2.

Now, let's consider each factor:

x or x+1 or x+7 or x+n must be a perfect square.

Each of these factors must be relatively prime, so if any two of them are perfect squares, they must differ by a perfect square.
We'll handle these cases one by one:

Case 1: x is a perfect square.

In this case, x=k^2. Then, x+1, x+7and x+n are consecutive integers starting from k^2. One of these consecutive integers must be a perfect square. Since consecutive perfect squares differ by at least 2k+1, the difference between two consecutive perfect squares must be less than or equal to 6 (since the largest difference here is x+7−x=7). This implies 2k+1≤6, so k≤2. Thus, we can check the values of k from 0 to 2.

Case 2: x+1 is a perfect square.

In this case, x+1=k^2, so x=k^2−1. Similarly x, x+7and x+n are consecutive integers starting from k2−1, and one of them must be a perfect square. Again, considering the difference between consecutive perfect squares, we get k≤2.

Case 3: x+7 is a perfect square.

In this case, x+7=k^2, so x=k^2−7. Following the same logic, we get k≤3.

Case 4: x+n is a perfect square.

This case is more general. We need to consider the possible values of n and k such that x+n=k^2. The difference between consecutive perfect squares limits the possibilities for n and k.

We'll now systematically check each of these cases to find all integer solutions (x, y) to the equation.

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