Question

Find two consecutive odd integers such that their product is 59 more than 2 times their sum.

Answer 1

so hmmm if you pick any integer, and multiply it by 2, you'd end up with an EVEN integer, 3 *2 = 6, 17 * 2 = 34 or, say 4 * 2 = 8 and so on

anyhow... to get an ODD one, simply add 1

so, hmmm let's use a reference integer, say "a"

so, our first odd number is then, ( 2a + 1)
now, to get our second number, we can simply add 2 to that, since
adding 2 to any odd integer, gives you another odd one, 3 +2 = 5
17 + 2 = 19 and so on

so our second number is ( 2a + 1) + 2
or just (2a + 3)

now, what's their product? well, (2a + 1)(2a+3)

what's their sum? well, (2a + 1) + (2a+3)

now two times that sum? well 2[ (2a + 1) + (2a+3) ]

now, 59 more than that? well 2[ (2a + 1) + (2a+3) ] + 5a

thus
`\bf (2a+1)(2a+3)=2[(2a+1)+(2a+3)]+59 \\\\\\ 4a^2+8a+3=2[4a+4]+59 \\\\\\ 4a^2+8a+3=8a+8+59\implies 4a^2+8a+3=8a+67 \\\\\\ 4a^2=67-3\implies 4a^2=64\implies a^2=\cfrac{64}{4}\implies a=√(16) \\\\\\ \boxed{a=4}`


now, since we know what "a" is, well, use that to get 2a+1 and 2a+3