Divisibility: How can I know if (5^12) -1 is divisible by 124?

[Helenam]

Not sure how to start looking at this. Any hints would be appreciated. Thanks
How can I know if (5^12) -1 is divisible by 124?

 

[Harry_the_cat]

Think of it as the difference of two squares and then again.

 

[Helenam]

Perfect. Thanks. (5^3 -1)

 

[Steven G]

It can be hard to decide if a SUM or DIFFERENCE can be divided by some number--like 124.
That is why you immediately think about factoring (unless the answer is obvious in the non factored form)!!

 

[Helenam]

I agree it’s not easy. I’m not sure I understand how to use factors. Could you show an example using factoring please.

 

[Dr.Peterson]

I agree it’s not easy. I’m not sure I understand how to use factors. Could you show an example using factoring please.

Isn't that what you did when you wrote this?

Perfect. Thanks. (5^3 -1)

That's a factor of 5^12 - 1, as a result of factoring the difference of squares twice. So you've done the factoring to solve the problem.

Is there something you are missing?

 

[khansaheb]

Not sure how to start looking at this. Any hints would be appreciated. Thanks
How can I know if (5^12) -1 is divisible by 124?

5¹² - 1 = 5¹² - 1¹² = (5^6 - 1) * (5^6 +1) = (5^3 - 1) * (5^3 + 1) * (5^6 + 1)

Do you understand the steps shown above? If not - please respond and tell us exactly where did you get lost?

 

[Steven G]

I agree it’s not easy. I’m not sure I understand how to use factors. Could you show an example using factoring please.

You want an example of factoring? Isn't that what you already did?

a^4 - b^4 = (a^2)^2 - (b^2)^2 = (a^2-b^2)(a^2 + b^2) = (a-b)(a+b)(a^2 + b^2)

 

[pka]

Not sure how to start looking at this. Any hints would be appreciated. Thanks
How can I know if (5^12) -1 is divisible by 124?

[imath] \left( {{5^{12}} - 1} \right) \\ = \left( {{5^6} - 1} \right)\left( {{5^6} + 1} \right) \\ =\left( {{5^3} - 1} \right)\left( {{5^3} + 1} \right)\left( {{5^6} + 1} \right) [/imath]
Now note that [imath]\bf{5^3-1=124}[/imath]

 

[Helenam]

Sorry it looks like I just got a bit confused. Yes I understood the factoring using the difference of two squares. I thought you meant there was some other method.

 

[Steven G]

Sorry it looks like I just got a bit confused. Yes I understood the factoring using the difference of two squares. I thought you meant there was some other method.

Just get the mess that you were given, factor it using any method of your choice, then hope that you can then see that 124 goes into the original number.
For the record, I do not see any way of factoring this other than by the difference of squares.

 

[BigBeachBanana]

[imath]5^3 \equiv 1 \mod 124\implies 5^{12} \equiv 1 \mod 124[/imath]

So we have that:
[imath]5^{12}-1 \equiv 0 \mod 124[/imath]

 

[Steven G]

[imath]5^3 \equiv 1 \mod 124\implies 5^{12} \equiv 1 \mod 124[/imath]

So we have that:
[imath]5^{12}-1 \equiv 0 \mod 124[/imath]

Yes! Thanks for showing this in the end. The thing is that beginning algebra students won't understand this.

 

[lookagain]

For the record, I do not see any way of factoring this other than by the difference of squares.

\(\displaystyle x^{4n} + x^{2n} + 1 \ = \)

\(\displaystyle (x^{4n} + 2x^{2n} + 1) \ - \ x^{2n} \ = \)

\(\displaystyle (x^{2n} + 1)^2 \ - \ x^{2n} \ = \)

\(\displaystyle (x^{2n} - x^n + 1)(x^{2n} + x^n + 1) \ \)

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\(\displaystyle 5^{12} - 1 \ = \)

\(\displaystyle (5^4 - 1)(5^8 + 5^4 + 1) \ = \)

\(\displaystyle (5^2 - 1)(5^2 + 1)(5^4 - 5^2 + 1)(5^4 + 5^2 + 1) \ =\)

\(\displaystyle (5^2 - 1)(5^2 + 1)(5^4 - 5^2 + 1)(5^2 - 5 + 1)(5^2 + 5 + 1) \ = \)

\(\displaystyle (24)(26)(601)(21)(31) \ = \)

\(\displaystyle (4)(31)[(6)(26)(601)(21)] \ = \)

\(\displaystyle 124[(6)(26)(601)(21)]\)