Number of roots

a 2 + b 2 + c 2 + d 2 + 12 < a b + 3 b + 6 c + 2 d \large { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+12<ab+3b+6c+2d

How many sets of 4 positive integers ( a , b , c , d ) (a,b,c,d) that satisfy the inequality above?

1 4 0 2

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2 solutions

a 2 + b 2 + c 2 + d 2 + 12 < a b + 3 b + 6 c + 2 d a 2 a b + b 2 3 b + c 2 6 c + d 2 2 d + 12 < 0 a 2 a b + b 2 4 + 3 4 ( b 2 4 b + 4 ) + c 2 6 c + 9 + d 2 2 d + 1 1 < 0 ( a b 2 ) 2 + 3 4 ( b 2 ) 2 + ( c 3 ) 2 + ( d 1 ) 2 < 1 \begin{aligned} a^2+b^2+c^2+d^2+12 & < ab + 3b+6c+2d \\ a^2-ab+b^2-3b+c^2-6c+d^2-2d+12 & < 0 \\ a^2-ab+\frac {b^2}4+\frac 34(b^2-4b+4)+c^2-6c+9+d^2-2d+1-1 & < 0 \\ \left(a-\frac b2\right)^2+\frac 34(b-2)^2+(c-3)^2+(d-1)^2 & < 1 \end{aligned}

We note that the LHS 0 \le 0 and equality occurs when a = 1 a=1 , b = 2 b=2 , c = 3 c=3 and d = 1 d=1 . Since the LHS is integral, other sets of ( a , b , c , d ) (a,b,c,d) will make it 1 \ge 1 . Therefore, there is only 1 \boxed{1} solution.

Linkin Duck
May 1, 2017

Since a , b , c , d a,b,c,d are all positive integers, we will find ( a , b , c , d ) (a,b,c,d) such that

a 2 + b 2 + c 2 + d 2 + 13 a b + 3 b + 6 c + 2 d a 2 + b 2 + c 2 + d 2 + 13 a b + 3 b + 6 c + 2 d 0 ( a 1 2 b ) 2 + 3 ( 1 2 b 1 ) 2 + ( c 3 ) 2 + ( d 1 ) 2 0 a = 1 , b = 2 , c = 3 , d = 1. { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+13\le ab+3b+6c+2d\\ \Longleftrightarrow { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+13-ab+3b+6c+2d\le 0\\ \Longleftrightarrow { \left( a-\frac { 1 }{ 2 } b \right) }^{ 2 }+3{ \left( \frac { 1 }{ 2 } b-1 \right) }^{ 2 }+{ \left( c-3 \right) }^{ 2 }+{ \left( d-1 \right) }^{ 2 }\le 0\\ \Longleftrightarrow a=1,b=2,c=3,d=1.

Hence, there is 1 \boxed { 1 } set ( a , b , c , d ) (a,b,c,d) .

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