# Iff - Interactive Proof Sorter (converse)

##### Age 14 to 18 Challenge Level:
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Hence, $8k+1=4n^2+4n+1$

Suppose $8k+1$ is a square number

Factorising the right hand side, $8k=4n(n+1)$

This is the general form for a triangular number. Therefore, if $8k+1$ is square, $k$ is triangular.

We wish to prove that if $8k+1$ is a square number then $k$ is a triangular number.

Simplifying, $k=\frac12n(n+1)$

All odd squares are of the form $(2n+1)^2=4n^2+4n+1$

Dividing both sides by $8$, $k=\frac{4n(n+1)}8$

Subtracting $1$ from both sides, $8k=4n^2+4n$

Since $8k$ is even, $8k+1$ is odd