Congruence in Z(integers) mod n

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Homework Statement



In Z mod 5, compute (a + b)5.


Homework Equations





The Attempt at a Solution



Noticing that (a+b)5 = a5+5(a4*b+2*a3*b2+2*a2*b3+a*b4)+b5. Since 5=0 in Z mod 5, it follows that 0(a4*b+2*a3*b2+2*a2*b3+a*b4)=0 and hence (a+b)5=a5+b5.

I am just wondering if it is correct for me to say 5=0 in Z mod 5 and just substitute 0 in for 5?
 
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\equivSounds reasonable to me. Why don't you test this with a couple of numbers to see if your results are consistent with what you've found?

BTW, instead of saying 5=0 in Z mod 5, you can say 5 \equiv 0 mod 5. That 3-bar equals sign means "is equivalent to".
 
thanks for the response, much appreciated
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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