Question

Let $T\left(n\right)$ denote the number of non-congruent triangles with integer side lengths and perimeter n. Thus $T\left(1\right)=T\left(2\right)=T\left(3\right)=T\left(4\right)=0$, while $T\left(5\right)=1$.Prove that -
i) $T\left(2006\right)<T\left(2009\right)$
ii) $T\left(2005\right)=T\left(2008\right)$

Answer 1

Let $a,b,c$ are the sides of triangle

Since triangle is non-congurent $a,b,c\epsilon I$ and different

Since n denotes the perimeter of triangle

$a+b+c=n$

$Case1:$ When n is odd perimeter is $\left[\frac{(n+3{)}^2}{48}\right]$ where [.] denotes greatest interger function

$Case2:$ When n is even perimeter is $\left[\frac{(n{)}^2}{48}\right]$ where [.] denotes greatest interger function

$\left(i\right)T\left(2006\right)=\left[\frac{(2006{)}^2}{48}\right]$

and $T\left(2009\right)=\left[\frac{(2012{)}^2}{48}\right]$

Here $\left[\frac{(2006{)}^2}{48}\right]<\left[\frac{(2012{)}^2}{48}\right]$

Hence $T\left(2006\right)<T\left(2009\right)$

$\left(ii\right)T\left(2005\right)=\left[\frac{(2008{)}^2}{48}\right]$

and $T\left(2008\right)=\left[\frac{(2008{)}^2}{48}\right]$

Here $\left[\frac{(2008{)}^2}{48}\right]=\left[\frac{(2008{)}^2}{48}\right]$

Hence $T\left(2005\right)=T\left(2008\right)$