three consecutive off integers are such that the square of the third integer is 105 less than the sum of the squares of the first two. one solution is -9,-7, and -5. find three other consecutive odd integers that also satisfy the given conditions

## -9,-7,and-5

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# 1 Answer

ok, the first thing we want to do is put this riddle is layman's terms. to start with i assume that's a typo up there and that you meant consecutive ODD integers. so we have 3 straight odd numbers ( like 1 3 5, 5 7 9, 9 11 13, etc...u get the idea). we
notice the difference between one odd number and the next is 2. (1, then 1+2 =3, and then 1+4 =5). the best way to define an odd number is in terms of integer numbers. let's say z is any old integer (like -23, -2, 0, 8, 77,...), now we can represent all the
odd numbers as 2z+1 !!! (try it, put any integer there for z and you will end up with an odd number) so now that we know we can call the first of these three numbers 2z+1, the next two numbers will be 2z+3, and 2z+5. the problem says if you square the 3rd
one, it's exactly 105 less than the sum of the squares of the first 2. this means: (2z+5)^2 = (2z+1)^2 + (2z+3)^2 - 105 now solve for z and you get the answer! okay, i'll walk you through it... perform all the squaring, this can get a little messy, but hang
in there: (4z^2+20z+25) = (4z^2+4z+1) + (4z^2+12z+9) - 105 4z^2+20z+25 = 4z^2+4z+1 + 4z^2+12z+9 - 105 (we really don't need those pesky parentheses) combine like terms... 4z^2+20z+25 = 8z^2+16z-95 now bring everybody to one side... (i choose the right side..doesn't
matter) 0 = 4z^2-4z-120 (that's just a quadratic equation!) 0 = z^2-z-30 (dividing both sides by 4..this just makes it easier to handle) 0 = (z-6)(z+5), solving for z we get -5 and 6 ...that means... 2z+1 = -9 (when z = -5) 2z+1 = 13 (when z = 6) so the triple
is really 13, 15, 17 since we already have -9, -7, -5!!!