<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] That's a helpful start.
Notice $(t-2)^5+(t+2)^5$ is an odd function:
$$ ((-t)-2)^5+((-t)+2)^5 = -(t+2)^5-(t-2)^5 = -\left((t-2)^5+(t+2)^5\right) $$
Therefore $A = \frac{(t-2)^{5}+(t+2)^{5}}{2t}$ is an even function, so when it's multiplied out and collected to a basic polynomial form, that form must be $A = Bt^4+Ct^2+D$. This is encouragement that it won't be that bad, plus it means we'll be able to finish the problem by considering $A-121$ as a quadratic in $t^2$.
I get $$ A = t^4+40t^2+80 $$ so the equation is
$$ 2t(t^4+40t^2+80) = 242 t$$
$$ 2t(t^4+40t^2-41) = 0 $$
$$ t(t^2-1)(t^2+41) = 0 $$
The solutions (three real and one complex pair) are $t \in \{-1, 0, 1, i\sqrt{41}, -i\sqrt{41}\}$, $x \in \{-2,-1,0,-1+i\sqrt{41},-1-i\sqrt{41}\}$.
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