Is $8r+1$ always a square for integer $r$?

Question

Assume that $r$ is an integer.

Since either $t$ or $t+1$ is even, $t$ is an integer for any integer $r$.

$$ \begin{align} 2r &= t(t+1)\\ 8r &= 4t(t+1)\\ 8r &= 4t^2 + 4t\\ 8r + 1 &= 4t^2 + 4t + 1\\ 8r+1 &= (2t+1)^2 \end{align} $$

So $8r+1$ must be an odd square for any integer $r$, however if $r=20$

$$ \begin{align} 8r+1&=8\cdot20+1\\ &=161\\ \sqrt{8r+1}&=\sqrt{161}\\ &\approx12.6886 \end{align} $$

What did I do wrong here? And when is $8r+1$ really a square number?

Answer 1

What is true is that every odd square is of the form $8r+1$. You can see this as follows:

$$(2n+1)^2=4n^2+4n+1=4n(n+1)+1$$ And $n(n+1)$ will always be even because it is the product of two consecutive integers, one of which will be even.

However, your argument goes in reverse and assumes that if you start with $r$ you can find an integer $t$ such that $r=\cfrac {t(t+1)}2$. The only positive integers $r$ of this form are triangular numbers $1,3,6,10 \dots$ so your initial assertion is false.

Answer 2

Any even number can't be written as the product of two consecutive number.

Hence in general, ∀r,∃t,2r=t(t+1) does not hold.

However, if for a given $r$, $∃t,2r=t(t+1)$ then your proof holds.

Example : $t=17\Rightarrow r=153$ and $8r+1=1225=(35)^2$