here's a nice one i came across while reading a book yesterday
prove for all positive integers m, the integer next greater than (\sqrt{3}+1)^{2m} contains
2^{m+1} as a factor
62This is in a way a standard question
Hint: you have to look at the expansion of
(√3-1)2m =f
1let I+f = (√3+1)2m
f' = (√3-1)2m
where 0<f,f'<1
adding,
I+f+f' = 2(C0+C23+C432+...+C2m3m)
now f+f' = integer
& 0<f+f'<2
=> f+f' = 1
so, int next greater than (√3+1)2m= 2(C0+C23+C432+...+C2m3m)
now how to show that 2m is a factor of (C0+C23+C432+...+C2m3m)
1Dimensions has already shown that the required integer is :
(\sqrt{3}+1)^{2m}+(\sqrt{3}-1)^{2m}=(4+2\sqrt{3})^{m}+(4-2\sqrt{3})^{m}
=2^{m}((2+\sqrt{3})^{m}+(2-\sqrt{3})^{m})
Now,
2|((2+\sqrt{3})^{m}+(2-\sqrt{3})^{m})
so, we can say that 2^{m+1} is a factor.