Let a, b, and c be positive integers with a b c such that a^2 - b^2

Question

Let a, b, and c be positive integers with a> b> c such that  a^2 - b^2 - c^2 + ab = 2011  and  a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997 . What is a?

(A) 249
(B) 250
(C) 251
(D) 252
(E) 253

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chetan2u · Accepted Answer

Let a, b, and c be positive integers with a> b> c such that  a^2 - b^2 - c^2 + ab = 2011  and  a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997 . What is a?

(A) 249
(B) 250
(C) 251
(D) 252
(E) 253

a^2 - b^2 - c^2 + ab = 2011  and 
 a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997 
Add the two equations to get  2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14 
 a^2 + b^2-2ab + a^2+c^2 - 2ac+b^2+c^2 - 2bc = 14 
 (a-b)^2+(a-c)^2+(b-c)^2=14

So 14 is sum of three squares=> Only possibility is 1+4+9
Thus, a-c=3, a-b=2 and b-c=1

Substitute c=a-3 and b=a-2 in  a^2 - b^2 - c^2 + ab = 2011 
 a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011 
 a^2-a^2-4+4a-a^2-9+6a+a^2-2a=2011.......8a=2011+13=2024...a=253

umangshah · Answer

Hi  Bunuel , can you share the answer?

andreagonzalez2k · Answer

Adding the two equations:
 a^2 - b^2 - c^2 + ab = 2011 
 a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997

we have:
 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14 
 a^2 + b^2 + c^2 - ab - ac - bc = 7 
 a^2 - a(b + c) + b^2 + c^2 - bc - 7 = 0

Solving for a:
 a = \frac{(b + c) +- \sqrt{(b + c)^2 - 4 (b^2 + c^2 - bc - 7)}}{2} =

=\frac{b + c +- \sqrt{b^2 + c^2 + 2bc - 4b^2 - 4c^2 + 4bc + 28}}{2} =

=\frac{b + c +- \sqrt{28 - 3b^2 - 3c^2 + 6bc}}{2} =

= \frac{b + c +- \sqrt{28 - 3(b - c)^2}}{2}

So we have two options:
1) b - c = 3
b = c + 3 -> a = c + 2 or a = c + 1
Not possible as a > b > c

2) b - c = 1
b = c + 1 -> a = c + 3 or a = c - 2
The only possible solution is a = c + 3

So a = c + 3 and b = c + 1

Substituting in the first equation:
 a^2 - b^2 - c^2 + ab = 2011

c^2 + 9 + 6c - c^2 - 1 - 2c - c^2 + c^2 + 4c + 3 = 2011

8c = 2000

c = 250

a = 250 + 3 = 253

bumpbot · Answer

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Let a, b, and c be positive integers with a> b> c such that a^2 - b^2

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