Question 243740
For most problems like this, your best interest is served by using as few variables as possible.  For example, the setup says "3 consecutive EVEN integers," which is a temptation to call x, y, and z.
But from what we know about consecutive numbers, they're just 1 apart.  Consecutive EVEN integers are 2 apart.  So we can define the 3 unknown numbers in terms of x.
a = x
b = x+2
c = x+4
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We're all told the square of the largest number {{{(x+4)^2}}} is 12 more than the sum of the squares of the other two numbers.  We can write this fact as: {{{x^2 + (x+2)^2 + 12}}}. Combining we have:
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{{{(x+4)^2 = x^2 + (x+2)^2 + 12}}}
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Resolving the squares.
{{{(x+4)(x+4) = x^2 + (x+2)(x+2) + 12}}}
{{{x^2 + 4x + 4x + 16 = x^2 + x^2 + 2x + 2x + 4 + 12}}}
{{{x^2 + 8x + 16 = 2x^2 + 4x + 16}}}
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Collecting like terms
Subtract x^2 from both sides
{{{8x + 16 = x^2 + 4x + 16}}}
Subtracting 8x from both sides
{{{16 = x^2 - 4x = 16}}}
Subtracting 16 from both sides
{{{0 = x^2 - 4x}}}
{{{x^2 - 4x = 0}}}
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We can factor an 'x' from the left-hand.
{{{x(x-4)=0}}}
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So we have two possible solutions:
x=0
x=4
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Testing x=0.
0^2 = 0
2^2 = 4
4^2 = 16
Is 4^2 equal to the sum of the other two squares + 12?
Yes.
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Testing x=4.
4^2 = 16
6^2 = 36
8^2 = 64
Is 8^2 equal to the sum of the other two squares + 12?
16+36=52, with is 12 less than 64.
So this works, too.
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The EVEN consecutive digits are:
0, 2 & 4
and
4, 6 & 8
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Done.