Prime, Prime, Prime
Considering the equations above with integers , what is the value of ?
The answer is 107.
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1 solution
Multiply all three equations by a factor of 2
Then we have
2a^2 + 2bc = 134
2b^2 + 2ac = 142
2c^2 + 2ab = 158
Add and you get
(a + b)^2 + (a + c)^2 + (b+c)^2 = 434
We know from inspection that 434 is the sum of the square of three consecutive numbers
Which numbers are they?
Let x will be the first number,
Based on this, create the equation:
x^2 + (x+1) ^2 + (x+2)^2 = 434
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 434
Combine the terms and solve the the quadratic equation
3x^2 + 6x + 5 = 434
3x^2 + 6x - 429 = 0
divide by 3
x^2 + 2x - 143=0
Solve for x => x = 11, which is the positive solution for the quadractic
So the other two numbers are 12 and 13
So we have 11^2 + 12^2 + 13^2 = 434
and
a + b = 11; b + c = 12; c + a = 13 ---- (1st set of equations)
Which yields
2a + 2b + 2c = 36
Or
a + b + c = 18 -> (2)
From (1) and (2) a = 5, b = 6, c = 7
So ab + bc + ca = 30 + 42 + 35 = 107