Let $a,b\in\Bbb N$ and $a+b$ be an even number.
Assume $a^2 - b^2 - a$ is an exact square, say $c^2$.
Let $m = \frac {a+b}2$ and $n = \frac {a-b}2$.
Then,
$$(4m-1)(4n-1) = 4(4mn-m-n) + 1 = 4(a^2-b^2-a)+1=(2c)^2+1^2 $$
My 2 questions are:
- Why does $(2c)^2+1$ have a prime divisor of the form $4k-1$?
- Why does it follow from the Fermat's Little Theorem that $4k-1$ divides $2c$ and $1$(thus contradiction)?
Part (1) can be done by a descent type argument.
If a number $N$ has a factor $4k-1$ then either this is prime and we're done or it factorises into two factors $p$ and $q$ with $1 \lt (p, q)\lt N$.
Consider multiplication modulo 4 to get to $4k-1$, which is $3 \mbox{ mod } 4$ we need one of $p, q$ is $3 \mbox{ mod } 4$ and the other is $1 \mbox{ mod } 4$.
WLOG take $p = 3 \mbox{ mod } 4 = 4j-1$ and we are back where we started but with a smaller integer. This process must terminate as the smallest positive integer of this form is $3$ itself.
Thus $2c^2+1$ has a prime divisor of the form $4k-1$.