Solve the equation $(5-x)^4+(2-x)^4=17$

Solution 1:

If you substitute $x= t+7/2$ the equation reduces to

$$2t^4+27t^2+ \frac{81}{8} = 17$$

which is quadratic in $t^2$.

The two ``obvious" solutions are equivalent to $t=1/2$ and $t=-1/2$ so you can factor the polynomial pretty quickly if you don't want to use quadratic formula.

(You get the $7/2$ by averaging $5$ and $2$.)

Solution 2:

I used hints in the comments to factorize the polynomial.

We know $x = 3$ or $x = 4$.

That is, $$ p(x) = (5-x)^4 + (2-x)^4 - 17 = 2(x-3)(x-4)(x-a)(x-b) $$

Where the factor $2$ is given by $x^4 + x^4$. To find $a$ and $b$, I put $x = 0$ and $x= 1$.

1. $x = 0$ gives $2\cdot 3 \cdot 4 ab = 5^4 - 1^4 = (5^2 + 1)(5^2 - 1) = 26\cdot 24$. We have $ab = 26$.

2. $x = 1$ gives $2\cdot 2 \cdot 3(1-a)(1-b) = 4^4 - 2^4 = (4^2 + 2^2)(4^2 - 2^2) = 240$. Therefore, $(1 - a)(1 - b) = 1 - a - b + ab = 20$, which gives $a + b = 7$.

So the resulting factorization is $$ p(x) = 2(x-3)(x-4)(x^2 - 7x + 26) $$

By solving $x^2 - 7x + 26 = 0$, we can find the remaining roots. I will leave it to you to find the roots, but it is obvious that they are not integers.