An algebra problem by Vishnu Kadiri

Algebra Level 3

Let P 1 = x 2 + a 1 x + b 1 P_1 = x^2 + a_1 x + b_1 and P 2 = x 2 + a 2 x + b 2 P_2 = x^2 + a_2 x + b_2 be polynomials with integer coefficients, and a 1 ≠ a 2 a_1 \ne a_2 . Suppose there exists some integers m m and n n satisfying P 1 ( m ) = P 2 ( n ) P_1(m) = P_2 (n) and P 1 ( n ) = P 2 ( m ) P_1 (n) = P_2 (m) , what can we conclude about the value of a 1 − a 2 a_1 - a_2 ?

Odd number Even number Irrational number Prime number

Vishnu Kadiri by Vishnu Kadiri , 19, India

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1 solution

Vishnu Kadiri
Jul 29, 2016

P 1 ( m ) = P 2 ( n ) = m 2 + a 1 m + b 1 = n 2 + a 2 n + b 2 → 1 P_1 (m)=P_2 (n)=m^2+a_1 m+b_1=n^2+a_2 n+b_2→1

P 1 ( n ) = P 2 ( m ) = n 2 + a 1 n + b 1 = m 2 + a 2 m + b 2 → 2 P_1 (n)=P_2 (m)=n^2+a_1 n+b_1=m^2+a_2 m+b_2→2 Subtracting 2 from 1, m 2 − n 2 + a 1 m − a 1 n = n 2 − m 2 + a 2 n − a 2 m m^2-n^2+a_1 m-a_1 n=n^2-m^2+a_2 n-a_2 m

( m + n ) ( m − n ) + a 1 ( m − n ) = ( n − m ) ( n + m ) + a 2 ( n − m ) (m+n)(m-n)+a_1 (m-n)=(n-m)(n+m)+a_2 (n-m)

( m − n ) ( m + n + a 1 ) = ( n − m ) ( n + m + a 2 ) (m-n)(m+n+a_1 )=(n-m)(n+m+a_2)

( m − n ) ( m + n + a 1 ) = ( m − n ) ∗ − 1 ( n + m + a 2 ) (m-n)(m+n+a_1 )=(m-n)*-1(n+m+a_2)

m + n + a 1 = − n − m − a 2 m+n+a_1=-n-m-a_2

m + n + a 1 − a 2 + a 2 = − n − m − a 2 m+n+a_1-a_2+a_2=-n-m-a_2

a 1 − a 2 = − m − n − a 2 − m − n − a 2 a_1 -a_2=-m-n-a_2-m-n-a_2

a 1 − a 2 = 2 ( − m − n − a 2 ) a_1 -a_2=2(-m-n-a_2) If some integer is multiplied by 2, then the product is even. ∴ a 1 − a 2 = 2 ( − m − n − a 2 ) ⇒ e v e n ∴a_1-a_2=2(-m-n-a_2 )⇒even
Q.E.D

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