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Algebra Level 3

If x = p + q x = \sqrt{p} + \sqrt{q} is a root of x 4 + a x 3 + b x 2 + c x + d = 0 x^4 + ax^3 + bx^2 + cx + d = 0 , where p p and q q are prime numbers and a a , b b , c c , and d d are integers, and if ( a + b ) 2 − 4 ( c + d ) = 816 (a + b)^2 - 4(c + d) = 816 , find the value of a + b + c + d + p + q a + b + c + d + p + q .

Inspiration


The answer is 176.

David Vreken by David Vreken , 42, USA

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1 solution

Pi Han Goh
Nov 25, 2020

Repeated squaring of x x gives x 2 = p + q + 2 p q ⟹ ( x 2 − p − q ) 2 = 4 p q ⟹ x 4 + x 3 ( 0 ) + x 2 ( − 2 p − 2 q ) + x ( 0 ) + ( p − q ) 2 = 0 x^2 = p + q + 2\sqrt{pq} \quad \implies \quad (x^2 - p -q)^2 = 4pq \quad \implies \quad x^4 + x^3 (0) + x^2(-2p - 2q) + x(0) + (p-q)^2 = 0 Thus, ( a , b , c , d ) = ( 0 , − 2 p − 2 q , 0 , ( p − q ) 2 ) (a,b,c,d) = (0, -2p-2q, 0, (p-q)^2) Substituting these values into the given constraint ( a + b ) 2 − 4 ( c + d ) = 816 (a+b)^2 - 4(c+d) = 816 and simplify gives p q = 51 ⇔ ( p , q ) = ( 3 , 17 ) , ( 17 , 3 ) ⟹ ( a , b , c , d ) = ( 0 , − 40 , 0 , 196 ) . pq = 51\quad \Leftrightarrow \quad (p,q) = (3,17), (17,3) \quad \implies \quad (a,b,c,d) = (0, -40, 0, 196) . The answer is 0 + ( − 40 ) + 0 + 196 + 3 + 17 = 176 . 0 + (-40) + 0 + 196 + 3 + 17 = \boxed{176}.

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