Puzzles

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left \(a\), the square number at the top left \(b\), and the triangle number at the top right \(c\). Adding upwards, we find that \(a+b+45=106\) and \(a+141+c=198\); and so \(a+b=61\) and \(a+c=57\). The only two square numbers that add to 61 are 25 and 36. Therefore \(c\) must be 21 or 32, but must be 21 as 32 is not a triangle number. And so \(a\) is 36 and \(b\) is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number:

Let \(T_a\) represent the \(a\)^th triangle number. This means that \(T_a=\frac{1}{2}a(a+1)\).

Suppose that for some integer \(n\), \(n^2 \leq T_a <(n+1)^2\). This means that:

But for every positive integer \(a \leq a^2\), so:

\(n\) and \(a\) are both positive integers, so:

Now consider \(T_{a+2}\):

We know that \(a \geq n\) and \(T_a \geq n^2\), so:

And so \(T_{a+2}\) is not between \(n^2\) and \((n+1)^2\). So if a triangle number \(T_a\) is between \(n^2\) and \((n+1)^2\) then the next but one triangle number \(T_{a+2}\) cannot also be between \(n^2\) and \((n+1)^2\). So there cannot be more than two triangle numbers between \(n^2\) and \((n+1)^2\).

Given an integer \(n\), how many triangle numbers are there between \(n^2\) and \((n+1)^2\)?

\(T_{n} = \frac{1}{2}n(n+1)\), so:

So, we are looking for \(n\) such that \((n+1)^2+(n+3)^2=(n+5)^2\). This is true when \(n=5\) (\(6^2+8^2=10^2\)).

Find \(n\) such that \(T_{n}+T_{n+1}+T_{n+1}+T_{n+2}=T_{n+2}+T_{n+3}\).