# A Quartic Equation

### Problem

Solve $(x+1)^{4}+(x+5)^{4}=82.$

### The way you've been taught

Being handed down an equation with integer coefficients of degree greater than 1, there is always a hope that the equation has integer solutions. If it does, they can be found via Viète's formulas, assisted by some guessing, division of polynomials, and good luck.

A *quartic* - fourth degree polynomial - with roots $\alpha,\beta,\gamma,\delta$ equals $a(x-\alpha)(x-\beta)(x-\gamma)(x-\delta),$ for some $a\ne 0.\;$ It follows that its free (of $x)$ coefficient equals $a\alpha\beta\gamma\delta,$ so that if it's an integer then the roots may be found among its divisors.

By direct verification,

$P(x)=(x+1)^{4}+(x+5)^{4}-82 = 2x^{4}+24x^{3}+156x^{2}+504x+544=0.$

$544=2^{5}\cdot 17.\;$ Since all coefficients of the equation are positive, it can't have positive roots.

Start checking:

$\begin{align} P(-1)&=0^{4}+4^{4}=256\ne 82,\\ P(-2)&=(-1)^{4}+3^{4}=1+81=82\\ P(-4)&=(-3)^{4}+1^{4}=81+1=82\\ P(-8)&=(-7)^{4}+(-3)^{4}\gt 82. \end{align}$

It is clear that, for other possible integer values of $x,$ $-16,-32,$ and the multiples of $17$ the polynomial exceeds $82.\;$ So there are two integer solutions $x=-2$ and $x=-4.\;$ A quadratic polynomial with these roots equals $Q(x)=x^{2}+6x+8,$ implying $P(X)=Q(x)R(x),$ where $R(x)$ is another quadratic polynomial, which can be found by the "long division algorithm" to be $Q(x)=2x^{2}+12x+68=2(x^{2}+6x+34).\;$ We may now safely apply the quadratic formula to solving $x^{2}+6x+34=0:$

$x_{1,2}=-3\pm\sqrt{9-34}=-3\pm 5i.$

where $i^{2}=-1.\;$ The equation thus has four roots: $-2,-4,-3\pm 5i.$

### Clever substitution 1

Set $a=x+1$ and $b=x+5.\;$ So defined $a$ and $b$ satisfy a system of equations:

$ a^{4}+b^{4}=82\\ b - a = 4. $

From the second equation, $a^{2}+b^{2}=(a-b)^{2}+2ab=2(8 + ab).\;$ In the same vein,

$\begin{align} a^{4}+b^{4} &= (a^{2}+b^{2})^{2}-2(ab)^{2}\\ &=4(8+ab)^{2}-2(ab)^{2}\\ &=256+64ab+2(ab)^{2}, \end{align}$

which leads to the equation $(ab)^{2}+32(ab)+87=0.\;$ With the quadratic formula, $(ab)_{1,2}=-16\pm\sqrt{256-87}=-16\pm 13;$ so that there are two possible values for the product $ab=(x+1)(x+5):$ $-3$ and $-29.\;$ The quadratic equation $(x+1)(x+5)=-3$ has two roots $-2$ and $-4.\;$ The second equation $(x+1)(x+5)=-29$ gives the complex pair $-3\pm 5i.$

### Clever substitution 2

Let $y=x+3.\;$ Then $x+1=y-2$ and $x+5=y+2.\;$ In terms of $y$ the equation becomes $(y-2)^{4}+(y+2)^{4}=82.\;$ From the binomial theorem, the odd powers of $y$ cancel out, leaving $y^{4}+24y^{2}-25=0.\;$ This is an example of *biquadratic*. Setting, say, $z = y^2$ reduces it to a quadratic equation in $z,$ which has two solutions $1$ and $-25.\;$ $y^{2}=1$ gives $y_{1,2}=\pm 1,$ so that $x+2=\pm 1.\;$ $y^{2}=25$ leads to two complex values, as above.

### Acknowledgment

The problem has been posted at the Short Mathematical Idea facebook group. The first clever solution is by Kunihiko Chikaya, the second by Regragui El Khammal.

|Up| |Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

62041991 |