MHB Is the Product \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) Divisible by 7?

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The product \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) is examined for divisibility by 7 for integers a, b, and c. The factors \(a^3-b^3\), \(b^3-c^3\), and \(c^3-a^3\) are analyzed, revealing that at least one of these differences must be divisible by 7 due to the properties of cubic residues modulo 7. Additionally, the term \(abc\) ensures that if any of a, b, or c is divisible by 7, the entire product is also divisible by 7. The conclusion is that the product is indeed divisible by 7 for any integers a, b, and c. Thus, the assertion holds true under the given conditions.
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Show that for integer a,b,c the product $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is divisible by 7
 
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kaliprasad said:
Show that for integer a,b,c the product $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is divisible by 7
If $a$, $b$, or $c$ is divisible by $7$, we are done.

Otherwise, each of $a^3$, $b^3$, $c^3$ is congruent to $\pm1\pmod{7}$. Therefore, at least two of these integers are congruent $\pmod7$, and the corresponding factor is divisible by $7$.
 

Thread 'Area of Overlapping Squares'

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