We had some great solutions to this problem, including both algebraic and geometric approaches.

Navjot from Sherborne Qatar sent us a diagram showing why the triangular numbers can be expressed using the formula $T_n=\frac12 n(n+1)$:

Rishika from Nonsuch High and Navjot both showed how to calculate the quadratic sequence by using differences. Here is Rishika's method:

Triangular number sequence: 1 3 6 10…

I noticed that this was a quadratic sequence, with the first differences

being 1,2,3,4.. and the second common difference being 1.

1 3 6 10 a+b+c

2 3 4 3a+b

1 1 2a

I then used the above expressions to solve and obtain the quadratic sequence rule:

$2a=1 \Rightarrow a=0.5$

$3(0.5) +b = 2$

$1.5 +b = 2$

$b = 0.5$

$a+b+c = 1$

$0.5+0.5+c = 1$

$c = 0$

Therefore, the $k^{th}$ term of the triangular number sequence is $0.5k^2 +0.5k$

Rishika, Navjot, Emily from South Hunsley and Samuel from West Bridgnorth all showed that if you multiply this by 8 and add 1, you always get a square number. Here's how Samuel worked it out:

$T_n = \frac{n^2+n}2$

$8T_n = 4n^2+4n$

$8T_n + 1 = 4n^2+4n+1$

This factorises to give $(2n+1)^2$

Eleanor from North Yorkshire gave a clear explanation for the second part of the problem:

Conjecture: If $8n+1 = m^2$ then $n$ is triangular.

Proof:

$8n+1 = 2(4n)+1$ so $8n+1$ must be odd as it is in the form $2(n)+1$

So if $8n+1$ is square then $8n+1 = (2k+1)^2$ since all odd squares have odd

roots

Then $8n+1 = 4k^2+4k+ 1$

So

$$\eqalign{n &= \frac{4k^2+4k}8 \\ &=\frac12(k^2+k) \\ &= \frac12k(k+1) \\ &= \sum_{r=1}^k r}$$

Hence if $8n+1 = m^2$ then $n$ is triangular.

Eleanor went on to show how to quickly check which numbers were triangular, by taking the square root of $8x+1$ for each value of $x$:

$\sqrt{8\times6214+1} \approx 222.9641227$ so $6214$ is not triangular.

$\sqrt{8\times3655+1} = 171$ so $3655$ is triangular.

$\sqrt{8\times7626+1} = 247$ so $7626$ is triangular.

$\sqrt{8\times8656+1} \approx 263.1520473$ so $8656$ is not triangular.

Pablo from King's College Alicante sent a diagram to show a geometric way of explaining what happens.

Pablo's Solution

Triangular numbers can be represented as a right angled triangle. For the

third triangular number the visual representation would have a base

of three, with two on top of it and one at the top (similar to the triangle in

Figure 2).

If 8 copies of one of these triangles are made, they can be arranged as

seen in Figures 1 and 2 to make an incomplete square. The only thing

missing would be an extra dot. Thus, if you take 8 copies of a Triangular

Number, plus an extra dot, a square can be made.

We also received solutions from Kristian from Maidstone Grammar, and Amrit from Hymers College. Thanks to you all!

Navjot from Sherborne Qatar sent us a diagram showing why the triangular numbers can be expressed using the formula $T_n=\frac12 n(n+1)$:

Rishika from Nonsuch High and Navjot both showed how to calculate the quadratic sequence by using differences. Here is Rishika's method:

Triangular number sequence: 1 3 6 10…

I noticed that this was a quadratic sequence, with the first differences

being 1,2,3,4.. and the second common difference being 1.

1 3 6 10 a+b+c

2 3 4 3a+b

1 1 2a

I then used the above expressions to solve and obtain the quadratic sequence rule:

$2a=1 \Rightarrow a=0.5$

$3(0.5) +b = 2$

$1.5 +b = 2$

$b = 0.5$

$a+b+c = 1$

$0.5+0.5+c = 1$

$c = 0$

Therefore, the $k^{th}$ term of the triangular number sequence is $0.5k^2 +0.5k$

Rishika, Navjot, Emily from South Hunsley and Samuel from West Bridgnorth all showed that if you multiply this by 8 and add 1, you always get a square number. Here's how Samuel worked it out:

$T_n = \frac{n^2+n}2$

$8T_n = 4n^2+4n$

$8T_n + 1 = 4n^2+4n+1$

This factorises to give $(2n+1)^2$

Eleanor from North Yorkshire gave a clear explanation for the second part of the problem:

Conjecture: If $8n+1 = m^2$ then $n$ is triangular.

Proof:

$8n+1 = 2(4n)+1$ so $8n+1$ must be odd as it is in the form $2(n)+1$

So if $8n+1$ is square then $8n+1 = (2k+1)^2$ since all odd squares have odd

roots

Then $8n+1 = 4k^2+4k+ 1$

So

$$\eqalign{n &= \frac{4k^2+4k}8 \\ &=\frac12(k^2+k) \\ &= \frac12k(k+1) \\ &= \sum_{r=1}^k r}$$

Hence if $8n+1 = m^2$ then $n$ is triangular.

Eleanor went on to show how to quickly check which numbers were triangular, by taking the square root of $8x+1$ for each value of $x$:

$\sqrt{8\times6214+1} \approx 222.9641227$ so $6214$ is not triangular.

$\sqrt{8\times3655+1} = 171$ so $3655$ is triangular.

$\sqrt{8\times7626+1} = 247$ so $7626$ is triangular.

$\sqrt{8\times8656+1} \approx 263.1520473$ so $8656$ is not triangular.

Pablo from King's College Alicante sent a diagram to show a geometric way of explaining what happens.

Pablo's Solution

Triangular numbers can be represented as a right angled triangle. For the

third triangular number the visual representation would have a base

of three, with two on top of it and one at the top (similar to the triangle in

Figure 2).

If 8 copies of one of these triangles are made, they can be arranged as

seen in Figures 1 and 2 to make an incomplete square. The only thing

missing would be an extra dot. Thus, if you take 8 copies of a Triangular

Number, plus an extra dot, a square can be made.

We also received solutions from Kristian from Maidstone Grammar, and Amrit from Hymers College. Thanks to you all!