Question 265347
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The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. 
Find the numbers.
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        The solution in the post by @mananth is incorrect. His answer is incorrect, as well.

        See below my correct solution.



<pre>
Let the three consecutive odd integer numbers be x, (x+2) and (x+4).


The equation is

    x^2 = 6(x + (x+2)) + 9,


Write it in standard form quadratic equation

    x^2 - 12x - 21 = 0.    (1)


Its discriminant is

    d = b^2 - 4ac = (-12)^2 -4*1*(-21) = 144 + 84 = 228.


The number 228 is not a perfect square.

It tells that equation (1) has no solutions in integer numbers.
</pre>

The conclusion is that the given problem has no solutions and describes a situation which NEVER may happen.


In other words, the problem as it is given in the post is a FAKE.