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Looks like Cauchy (Posted on 2019-04-17) Difficulty: 3 of 5
For all positive integers k, define f(k)=k2+k+1. Compute the largest positive integer n such that

2019f(12)f(22)...f(n2)>=(f(1)f(2)...f(n))2

No Solution Yet Submitted by Danish Ahmed Khan    
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Better solution Comment 3 of 3 |
The key to the problem is the fact f(k-1)*f(k)=f(k^2)
(k^2-k+1)(k^2+k+1)=k^4+k^2+1

Which means
f(k)^2/f(k^2)=f(k)/f(k-1)

So if we rewrite the inequality in the problem as

2019 >= f(1)^2/f(1^2) * f(2)^2/f(2^2) *... * f(n)^2/f(n^2)

Each individual term reduces

2019 >= f(1)/f(0) * f(2)/f(1) * ...* f(n)/f(n-1)

And most of the terms cancel out leaving

2019 >= f(n)/f(0) = f(n)

2019 >= n^2 + n + 1

n <= 44


  Posted by Jer on 2019-04-19 10:33:24
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