IIT-JEE 2000Question 1
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.
Sequence and Series · Arithmetic Progression
Answerproved
Solution
Formulas used. (1) $a(a+3d)=a^2+3ad$ and $(a+d)(a+2d)=a^2+3ad+2d^2$(2) $x(x+2y)+y^2=(x+y)^2$Step 1 - Pair the four terms. For $a,\ a+d,\ a+2d,\ a+3d$, put $x=a^2+3ad$: $$a(a+3d)\cdot(a+d)(a+2d)=x\left(x+2d^2\right).$$ Step 2 - Add $d^4$. With $y=d^2$ in formula (2): $$x\left(x+2d^2\right)+d^4=\left(x+d^2\right)^2=\left(a^2+3ad+d^2\right)^2 .$$ Step 3. $a,d$ are integers, so $a^2+3ad+d^2$ is an integer, and the sum is the square of an integer. Answer: proved