Symmetry-Based Re-Parameterizations

Choose the Middle Term!

When looking at arithmetic sequences, why look at them as $a$, $a+d$, $a+2d$, $a+3d$, $a+4d$? Why not look at them as $a-2d$, $a-d$, $a$, $a+d$, $a+2d$? Symmetry makes nice cancellations occur!

Similarly, when looking at geometric sequences why look at them as $a$, $ar$, $ar^2$, $ar^3$, $ar^4$? Why not look at them as $\frac{a}{r^2}$, $\frac{a}r$, $a$, $ar$, $ar^2$?

Problem (ARML 2011): The product of the first five terms of a geometric sequence is $32$. If the fourth term is $17$, compute the second term.

Solution: Let the terms be $\frac{a}{r^2}$, $\frac{a}r$, $a$, $ar$, $ar^2$. Then $\frac{a}{r^2}\frac{a}r(a)(ar)(ar^2)=32$ so $a^5=32$ so $a=2$. Then $17=ar=2r$ so $r=\frac{17}2$. Therefore, the second term is just $\frac{2}{17/2}=\frac4{17}$.

Problem (ARML 2012): The four zeros of $x^4+jx^2+kx+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$.

Solution: This trick doesn't only apply to an odd number of terms! Take $a-\frac32d$, $a-\frac12d$, $a+\frac12d$, $a+\frac14d$. By Vieta's formulas, the roots sum to $0$ so $a=0$! Therefore the roots are just $-\frac32d$, $-\frac12d$, $\frac12d$, $\frac32d$, but their product must be

$$\frac{3^2}{2^4}d^4=225=15^2\implies d=\pm2\sqrt5$$

We can assume $d=2\sqrt5$ by symmetry, so the roots are $-3\sqrt5$, $-\sqrt5$, $\sqrt5$, and $3\sqrt5$. We can easily find $j$ from here.

Problem (104 Number Theory Problems by Titu Andreescu): Give an example of $11$ consecutive positive integers the sum of whose squares is a perfect square.

Solution: Consider $a-5d$, $a-4d$, $\dots$, $a+4d$, $a+5d$:

$$\begin{align*}\sum_{i=-5}^5(a+id)^2&=a^2+\sum_{i=1}^5[(a+id)^2+(a-id)^2]\\&=a^2+\sum_{i=1}^5[2a^2+2i^2d^2]\\&=11a^2+110d^2=11(a^2+10d^2)\end{align*}$$

So $11\mid a^2+10d^2$, therefore $a^2\equiv d^2\pmod{11}$ or $(a-d)(a+d)\equiv0\pmod{11}$. This simplifies the problem vastly.

This technique helps on Olympiads too! Don't believe me?

Problem (Dutch Mathematical Olympiad 1998): Find all real solutions to the following equation:

$$(x+1995)(x+1997)(x+1999)(x+2001)+16=0$$

Solution: Take $u=x+1998$!

$$0=(u-3)(u-1)(u+1)(u+3)+16=(u^2-1)(u^2-9)+16$$

which is a quadratic in $u^2$. We can solve this easily!