<s>[INST] Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
 [/INST] The “brute-force” way is to use the Binomial Theorem:
$$(x+a)^5 = x^5 + 5ax^4 + 10a^2x^3 + 10a^3x^2 + 5a^4x + a^5$$
Which gives us:
$$(x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1) + (x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243) = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 410x + 242 = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x = 0$$
$$2x(x^4 + 5x^3 + 50x^2 + 130x + 84) = 0$$
Clearly, $x=0$ is a solution.  By the rational root theorem, roots of the quartic factor must be a factor of 84 (i.e., in $\pm\{ 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84\}$).  Descartes' rule of signs rules out all positive solutions.  Trial and error gives $x = -1$ and $x = -2$ as roots.
Dividing the quartic by the factors $(x+1)(x+2)$ gives the quadratic equation $x^2 + 2x + 42$, with the roots $-1 \pm i \sqrt{41}$.
Putting this all together, $x \in \{ 0, -1, -2, -1 + i\sqrt{41}, -1 - i\sqrt{41} \}$.
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