Diamond Numbers and Fermat's method of Factorization I call numbers which are the product of two consecutive numbers $n(n+1)$ "diamond numbers". Can they be used in factorizing numbers using Fermat's method of the difference of two squares,considering that $(n + \frac 1 2)^2 = n(n+1) + \frac 1 4$? I had in mind something like $(5+\frac12)^2 -(3+\frac12)^2 = 2\times9 =18$
 A: Such numbers $n(n+1)$ are even, so the differences between them are also even.  It turns out that this could be related to factorising even numbers into an even and an odd number, since  $$2a(2b+1) = \left(a+b\right)\left(a+b+1\right)- \left(\left|a-b-\frac12\right|-\frac12\right)\left(\left|a-b-\frac12\right|+\frac12\right) $$ $$c(c+1)-d(d+1)=(c+d+1)(c-d)$$ where the second right hand side is the product of numbers of different parity. So for example you have 


*

*$60 = 30\times 31 -29\times 30 = (30+29+1)(30-29)=60\times 1$

*$60 = 11\times 12 -8\times 9 = (11+8+1)(11-8)=20\times 3$

*$60 = 9\times 10 -5\times 6 = (9+5+1)(9-5)=15\times 4$

*$60 = 8\times 9 -3\times 4 = (8+3+1)(8-3)=12\times 5$


but there is an easier way of finding at least one such factorisation of an even number: keep dividing by $2$ until it stops being even. 
Added:
If you prefer to consider squares of numbers which are $\frac12$ more than integers, and their differences, this becomes $$ 2a(2b+1) = \left(a+b+\frac12\right)^2 - \left(a-b-\frac12\right)^2$$ $$\left(c+\dfrac12\right)^2-\left(d+\dfrac12\right)^2=(c+d+1)(c-d)$$ and 


*

*$60 = 30.5^2 - 29.5^2 = (30.5+29.5)(30.5-29.5) = 60\times 1$

*$60 = 11.5^2 -8.5^2 = (11.5+8.5)(11.5-8.5)=20\times 3$

*$60 = 9.5^2 -5.5^2 = (9.5+5.5)(9.5-5.5)=15\times 4$

*$60 = 8.5^2 -3.5^2 = (8.5+3.5)(8.5-3.5)=12\times 5$


essentially the same as my initial response. It is still the factorisation of even numbers into an even and an odd number.
