The fourth power of the common...

MathematicsArithmetic Progression (A.P.)JEE Advanced 2000Easy

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 (English)

Defining the Terms of the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A . P .

Let the four consecutive terms of the $A . P .$ be $x, x + D, x + 2 D, x + 3 D$ .
Here, $x$ is the first term and $D$ is the common difference.
Since the entries are integers, $x, D \in Z$ .

Setting up the Expression

We need to evaluate the sum $S = Product + D^{4}$ .
The expression is: $S = x (x + D) (x + 2 D) (x + 3 D) + D^{4}$ .

Strategic Grouping of Terms

Rearrange the terms to group the $1^{s t}$ with the $4^{t h}$ and the $2^{n d}$ with the $3^{r d}$ .
$S = [x (x + 3 D)] \cdot [(x + D) (x + 2 D)] + D^{4}$

Identifying the Common Quadratic Part

Expanding the groups:
$x (x + 3 D) = x^{2} + 3 x D$
$(x + D) (x + 2 D) = x^{2} + 3 x D + 2 D^{2}$
So, $S = (x^{2} + 3 x D) (x^{2} + 3 x D + 2 D^{2}) + D^{4}$

Substitution for Simplification

Let $y = x^{2} + 3 x D$ .
Substitute $y$ into the expression for $S$ :
$S = y (y + 2 D^{2}) + D^{4}$

Expanding and Recognizing the Identity

Expand the expression:
$S = y^{2} + 2 y D^{2} + D^{4}$
Notice that this is in the form $a^{2} + 2 ab + b^{2}$ where $a = y$ and $b = D^{2}$ .

Forming the Perfect Square

Using the identity $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$S = (y + D^{2})^{2}$

Conclusion and Final Proof

Substitute $y = x^{2} + 3 x D$ back into the expression:
$S = (x^{2} + 3 x D + D^{2})^{2}$
Since $x, D \in Z$ , the term $(x^{2} + 3 x D + D^{2})$ is also an integer.
Conclusion: The resulting sum is the square of an integer.

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