A awesome mathematics problem | Olympiad Question | can you solve this problem | x=?

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[@pietergeerkens6324]

I believe the following is easier to follow, and with much smaller magnitude constants makes the arithmetic and algebra simpler. This is a valuable consideration under exam or contest conditions, as increasing confidence that one's algebra is correct on the first pass.
Dividing through by 4 and letting u = √x / 2¹⁰ yields:
⁵√[ ½ + u ] + ⁵√[ ½ - u ] = 1
Then defining
a = ⁵√[ ½ + u ]
b = ⁵√[ ½ - u ]
v = ab = ⁵√[ ¼ - u² ]
we obtain
a + b = 1
ab = v
and
a² + b² = (a+b)² - 2ab = 1 - 2v
a³ + b³ = (a+b)³ - 3ab(a+b) = 1 - 3v
and finally that
1 = a⁵ + b⁵ = (a² + b²)(a³ + b³) - (ab)²(a+b) = 1 - 5v + 6v² - v²
or
5v² - 5v = 5⋅v⋅(v-1) 0.
Now v = 1 can be rejected as requiring u² < 0, and
v = 0
gives
u² = ¼
and finally
x = (2¹⁰)²⋅u²
= 2¹⁸.

[@RealQinnMalloryu4]

{x^5+256}= 256x^5+{x+x ➖ } =x^2 {256x^5+x^2}= 256x^7 10^20^7^8x^7 2^55^4^1^2^3x^1 1^1^12^21^1 1^2;(x ➖ 2x+1) (x^5)^2={x^25 ➖ 256 }= 231 (x)^2= x^2 {231 ➖ x^2}=229 10^20^3^2 2^55^43^2 1^1^12^23^2 1^2^3^1 23 (x ➖ 3x+2)