The fourth power of the common difference of an arithmetic progression with integer entries is added

Q: Why do we group the first and fourth terms? Is there a rule for this?

There is no strict rule, but it is a powerful heuristic in algebra. By pairing the terms with the smallest and largest values, we often create products that share a common quadratic part. In this case, $x(x+3D) = x^2 + 3xD$ and $(x+D)(x+2D) = x^2 + 3xD + 2D^2$. This commonality is the key to simplifying the expression.

Q: Can we just expand the whole thing and hope for the best?

You certainly could! If you expand $x(x+D)(x+2D)(x+3D) + D^4$ completely, you will get a quartic polynomial. While it will eventually factor into a perfect square, it is much more prone to calculation errors. The grouping method is not just faster; it is more elegant and less likely to lead to mistakes.

Q: How do we know the final expression is an integer?

The problem states that the entries of the arithmetic progression are integers, meaning $x$ and $D$ are integers. Since the set of integers is closed under addition and multiplication, any polynomial expression involving only integer coefficients and integer variables will also result in an integer. Thus, $(x^2 + 3xD + D^2)^2$ is the square of an integer.

Question

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

Sigma Percentile · Accepted Answer

### The Elegance of Arithmetic Progressions

My dear student, welcome to a journey through the heart of number theory and algebraic beauty. Today, we are not just solving a problem; we are uncovering a hidden symmetry in arithmetic progressions.

Imagine you are standing on a path where each step is defined by a constant difference. We are looking at four consecutive terms of an arithmetic progression, and we are asked to prove that adding the fourth power of the common difference to their product results in a perfect square.

It sounds daunting, but let us peel back the layers together.

### The Setup: Defining Our Universe

Let us start by defining our terms. We have an arithmetic progression with integer entries. Let the first term be $x$ and the common difference be $D$.

Our four consecutive terms are $x$, $x+D$, $x+2D$, and $x+3D$. Since the entries are integers, we know that $x, D \in \mathbb{Z}$. This integer constraint is our bedrock; it ensures that our final result will be a perfect square of an integer.

We are interested in the sum $S$, defined as the product of these four terms plus the fourth power of the common difference:

$$S = x(x+D)(x+2D)(x+3D) + D^4$$

If you were to multiply this out blindly, you would face a fourth-degree polynomial. It is not impossible, but it is a trap—a path filled with potential for arithmetic errors. Instead, let us look for the hidden symmetry.

### The Strategic Insight: Finding the Commonality

Look closely at the terms. If we pair the first term with the fourth, and the second with the third, something magical happens. Let us group them:

$$S = [x(x+3D)] \cdot [(x+D)(x+2D)] + D^4$$

Now, let us expand these pairs. The first pair, $x(x+3D)$, becomes $x^2 + 3xD$. The second pair, $(x+D)(x+2D)$, expands to $x^2 + 2xD + xD + 2D^2$, which simplifies to $x^2 + 3xD + 2D^2$.

Do you see it? Both groups contain the term $x^2 + 3xD$. This is the "Aha!" moment that turns a complex problem into a simple one.

### The Substitution: Simplifying the Complexity

To make our lives easier, let us perform a substitution. Let $y = x^2 + 3xD$.

Now, our expression for $S$ becomes incredibly clean:

$$S = y(y + 2D^2) + D^4$$

We have transformed a daunting product of four terms into a simple quadratic in terms of $y$. Let us expand this:

$$S = y^2 + 2yD^2 + D^4$$

### The Identity: The Final Reveal

Take a deep breath and look at that expression: $y^2 + 2yD^2 + D^4$. Does it look familiar?

It is the classic algebraic identity $a^2 + 2ab + b^2 = (a+b)^2$, where $a = y$ and $b = D^2$. We can compress this into:

$$S = (y + D^2)^2$$

We have successfully transformed that massive, intimidating product into a neat, compact perfect square.

### The Conclusion: A Mathematical Triumph

Finally, let us bring back our original variables. Substituting $y = x^2 + 3xD$ back into our result, we get:

$$S = (x^2 + 3xD + D^2)^2$$

Since $x$ and $D$ are integers, the expression $x^2 + 3xD + D^2$ is also an integer. Therefore, the sum $S$ is indeed the **square of an integer**.

We have not just solved the problem; we have proven a fundamental property of arithmetic progressions. Keep this spirit of curiosity alive, and remember: in mathematics, there is always a more elegant path if you look closely enough.

Animated Solution for Mathematics - Sequence and Series: The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

Visualized Solution

Defining the Terms of the A.P.

Setting up the Expression

Strategic Grouping of Terms

Identifying the Common Quadratic Part

Substitution for Simplification

Expanding and Recognizing the Identity

Forming the Perfect Square

Conclusion and Final Proof

The Sigma Insight: Arithmetic Progression (A.P.)

The Elegance of Arithmetic Progressions

The Setup

The Strategic Insight

The Substitution

The Identity

The Conclusion

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