Simply expand it and compute the discriminant

Algebra Level 2

( b + c − a ) ( a + b − c ) ( a − b + c ) ( a + b + c ) × ( a − b − c ) ( b − a − c ) ( c − a − b ) ( − a − b − c ) \large{\color{#624F41}{(\color{#3D99F6}{b}\color{#20A900}{+c}\color{#D61F06}{-a})(\color{#D61F06}{a}\color{#3D99F6}{+b}\color{#20A900}{-c})(\color{#D61F06}{a}\color{#3D99F6}{-b}\color{#20A900}{+c})(\color{#D61F06}{a}\color{#3D99F6}{+b}\color{#20A900}{+c})\\ \times (\color{#D61F06}{a}\color{#3D99F6}{-b}\color{#20A900}{-c})(\color{#3D99F6}{b}\color{#D61F06}{-a}\color{#20A900}{-c})(\color{#20A900}{c}\color{#D61F06}{-a}\color{#3D99F6}{-b})(\color{#D61F06}{-a}\color{#3D99F6}{-b}\color{#20A900}{-c})}}

Suppose a , b , c a,b,c are integers, is the above product a square?

No Yes

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Nihar Mahajan by Nihar Mahajan , 20, India

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1 solution

Nihar Mahajan
Oct 2, 2015

( − a + b + c ) ( a + b − c ) ( a − b + c ) ( a + b + c ) × ( a − b − c ) ( − a + b − c ) ( − a − b + c ) ( − a − b − c ) = ( − a + b + c ) ( a + b − c ) ( a − b + c ) ( a + b + c ) × ( − 1 ) ( − a + b + c ) ( − 1 ) ( a − b + c ) ( − 1 ) ( a + b − c ) ( − 1 ) ( a + b + c ) = ( − a + b + c ) ( − a + b + c ) ( a + b − c ) ( a + b − c ) ( a − b + c ) ( a − b + c ) ( a + b + c ) ( a + b + c ) ( − 1 ) 4 = ( − a + b + c ) 2 ( a + b − c ) 2 ( a − b + c ) 2 ( a + b + c ) 2 (-a+b+c)(a+b-c)(a-b+c)(a+b+c) \\ \times (a-b-c)(-a+b-c)(-a-b+c)(-a-b-c) \\ = (-a+b+c)(a+b-c)(a-b+c)(a+b+c) \\ \times(-1)(-a+b+c)(-1)(a-b+c)(-1)(a+b-c)(-1)(a+b+c) \\ = (-a+b+c)(-a+b+c)(a+b-c)(a+b-c)(a-b+c)(a-b+c)(a+b+c)(a+b+c)(-1)^4 \\ = (-a+b+c)^2(a+b-c)^2(a-b+c)^2(a+b+c)^2

Thus , the given expression is a perfect square.

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. +1

Sharky Kesa - 5 years, 8 months ago

Simple plug a = b = c = 1 a=b=c=1 (lol)

Mehul Arora - 5 years, 8 months ago

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