Question 1180329
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The good solution from tutor @ikleyn is a good example of how choosing the 3 consecutive even integers as x-2, x, and x+2 can make a problem much easier to solve using formal algebra than calling them x, x+2, and x+4.<br>
Keep that in mind when you solve other problems about consecutive integers.<br>
The same concept can be used to solve the problem informally in a very short time.<br>
The three integers are close together, so the sum of the squares of the three integers is close to 3 times the square of the middle one.<br>
1208 divided by 3 is a bit more than 400, and 400 is 20^2.<br>
So the numbers should be 18, 20, and 22.  Perform quick calculations to verify that 18^2+20^2+22^2=1208.<br>
Then remember that since the problem is about the sum of the squares of the three integers, another solution is -18, -20, and -22.<br>