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 Looks like Cauchy (Posted on 2019-04-17)
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 No Rating

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 Weak solution | Comment 2 of 3 |
f(k)^2=k^4+2k^3+3k+2+2k+1
f(k^2)=k^4+k^2+1

The ratio of these values forms a nice pattern:
k=1, 3/1
k=2, 7/3
k=3, 13/7
k=4, 21/13
k=5, 31/21

So the product of these is just the last numerator, which we need to be at most 2019.  Continuing the table

k=44, 1981/1893 (smaller)
k=45, 2071/1981 (too big)

Without looking too hard into why this works, the solution is n=44

 Posted by Jer on 2019-04-18 11:07:17

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