A Quartic Equation

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

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: $-2,-4,-3\pm 5i.$

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 $-3\pm 5i.$

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.

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.