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[@thelifeclass]

Interesting🎉

[@JotaAlbertoLD]

EXCELENT 🙌

[@johnstanley5692]

Use Zn := y^n + 1/y^n , problem is to find Zn=38. Here y = 10+sqrt(3) => Z(1) = 2*sqrt(10). Recursion Z(n+1)=Z(n)Z(1)-Z(n-1) or doubling
formula Z(2n) = Z(n)^2=2 . In this case Z(2)=Z(1)^2-2 = 38 => n = 2 (or x=2 in original notation)