Iff - Interactive Proof Sorter (converse)

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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