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Sagot :
2n+1, 2n+3 - consecutive odd integers
[tex](2n+1)(2n+3)=3(2n+1+2n+3)+6\\ 4n^2+6n+2n+3=3(4n+4)+6\\ 4n^2+8n+3=12n+12+6\\ 4n^2-4n+15=0\\ \Delta=(-4)^2-4\cdot4\cdot15=16-240=-224\\[/tex]
[tex]\Delta<0\Rightarrow n\in \emptyset [/tex]
It's not possible.
[tex](2n+1)(2n+3)=3(2n+1+2n+3)+6\\ 4n^2+6n+2n+3=3(4n+4)+6\\ 4n^2+8n+3=12n+12+6\\ 4n^2-4n+15=0\\ \Delta=(-4)^2-4\cdot4\cdot15=16-240=-224\\[/tex]
[tex]\Delta<0\Rightarrow n\in \emptyset [/tex]
It's not possible.
I don't think it is.
If the first integer is 'x', then the second one is (x+2).
Their product is (x²+2x), and their sum is (2x+2).
You have said that (x²+2x) = 3(2x+2) + 6
x²+2x = 6x + 6 + 6
x² - 4x - 12 = 0
(x - 6) (x + 2) = 0
x = 6 and x = -2
The numbers are ('6' and '8'), or ('-2' and 0).
I honestly don't know what to make of the second pair.
But the first pair satisfies the description:
The product (48) = 3 x the sum (3 x 14 = 42) plus 6.
So '6' and '8' completely work, except that they're not odd integers.
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