parmenides51 wrote:
Find the natural number n for which √20n−18n19is a rational number.
Translated from here:
**Solution.** n=0 is a solution
For n≥1, 20n−18n19 is a perfect square, 20n−18n=19x2, 2n(10n−9n)=19x2, x is a natural number; n is even, n=2m, m is a natural non-zero number
22m(102m−92m)=19x2, x is a natural non-zero number, x=2my, y is odd, 102m−92m=19y2,
(10m−9m)⋅(10m+9m)=19y2, (10m−9m,10m+9m)=1
**Case 1:** There exist odd natural numbers a,b, (a,b)=1, a⋅b=y with 10m+9m=19a2 and
10m−9m=b2; 10m=9m+b2=M4+1+M4+1=M4+2,
m=1,b=1,a=1,y=1,x=2,n=2
**Case 2:** There exist odd natural numbers a,b, (a,b)=1, a⋅b=y with 10m+9m=a2 and
10m−9m=19b2; 10m=9m+19b2=M8+1+19(M8+1)=M8+4,
m=2,b=1,a2=181, contradiction
S={0,2}