Two consecutive odd integers are [tex]2k+1[/tex] and [tex]2k+3[/tex], for some integer [tex]k[/tex].
Their product is [tex](2k+1)(2k+3)=4k^2+8k+3[/tex].
Twice their sum is [tex]2(2k+1+2k+3)=2(4k+4)=8k+8[/tex]
Since the product is 1 less than twice the sum, we have
[tex]\underbrace{4k^2+8k+3}_{\text{the product}}=\underbrace{8k+8}_{\text{twice the sum}}-1[/tex]
So, we have
[tex]4k^2-4=0 \iff 4k^2=4 \iff k^2=1 \iff k=\pm 1[/tex]
If [tex]k=1[/tex], the integers are 3 and 5
If [tex]k=-1[/tex], the integers are -1 and 1.
In both cases, in fact, we have:
