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

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Integer Product
kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
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
 
Mathematics news on Phys.org
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 $7$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top