A simple function

Algebra Level 5

Define a function $f:Z\rightarrow Z$ such that $f(x)={ x }^{ 2 }+x+1$ for every integer x. Find the largest positive integer "n" such that:

$2015\times f({ 1 }^{ 2 })\times f({ 2 }^{ 2 })⋯f({ n }^{ 2 })\quad ≥\quad { (f(1).f(2)⋯f(n)) }^{ 2 }$ .

1 solution

Gaurav Jain
Feb 3, 2015

Note the factorization.

$n^{ 4 }+n^{ 2 }+1= (n^{ 2 }+n+1)(n^{ 2 }-n+1)= (n^{ 2 }+n+1)((n-1)^{ 2 }+(n-1)+1)$ .

Hence: $f\left( x \right)$ satisfies the relation $f(n^{ 2 })=f(n)\times f(n-1)$ .

Dividing both sides by $∏(product) f(n2)$ .

On simplification it turns into

$2015 ≥\frac { f(n) }{ f(0) } = { n }^{ 2 }+n+1$ .

Now since

${ n }^{ 2 }<n^{ 2 }+n+1 < (n+1)^{ 2 }$ and $44^{ 2 } < 2015< 45^{ 2 }$

So ,It suffices to check only n=44, which still satisfies the inequality.

Creative solutions are heartedly invited.

Gaurav Jain - 6 years, 4 months ago