# Primality Proof

Prove or disprove that $$\large \frac{5^{125}-1}{5^{25}-1}$$ is prime.

This problem is not original

Note by Jihoon Kang
3 years, 9 months ago

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$\frac{x^5-1}{x-1} = \frac{(x-1)(x^4 +x^3 +x^2 +x +1)}{x-1} = 5^{100} +5^{75} +5^{50} +5^{25} +1 = y$ but $3597751 | y \therefore\ not \ prime$ You also might be able to use the pseudo prime test i.e. $a^{y-1} \neq 1 mod(y) \ for \ gcd(a,n) = 1$

- 3 years, 9 months ago

FYI For exponents with multiple digits, use { } for all of them to display properly, EG 5^{100}

How did you guess that $$3597751 \mid y$$? Or did you use a computer?

Staff - 3 years, 9 months ago

yea I used wolfram alpha

- 3 years, 9 months ago

Haha. I guess this is a valid proof :) I proved it using difference of two squares.

- 3 years, 9 months ago

Great approach! Can you share your solution?

Staff - 3 years, 9 months ago

Certainly

Let $$x=5^{25}$$

Then $$\frac{5^{125}-1}{5^{25}-1}=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$$

You can notice that $$x^4+x^3+x^2+x+1=(x^2+3x+1)^2-(5x^3+10x^2+5x)$$

$$=(x^2+3x+1)^2-5x(x^2+2x+1)$$

$$=(x^2+3x+1)^2-5^{26}(x+1)^2$$

And then, using difference of two squares, you get:

$$((x^2+3x+1)-5^{13}(x+1))((x^2+3x+1)+5^{13}(x+1))$$

So, the above expression has these two factors, hence it is not prime.

Q.E.D

- 3 years, 9 months ago

Cool proof. How'd you get $$x^4+x^3+x^2+x+1=(x^2+3x+1)^2-(5x^3+10x^2+5x)$$, though? Was it purely intuition?

- 3 years, 6 months ago

Yeah. I thought that difference of two squares may be the way to go, so I got $$(Ax^2+Bx+1)^2$$ and, through trial and error, found (fortunately) appropriate values for A and B so that it worked. It could have easily not worked either.

- 3 years, 5 months ago

- 3 years, 9 months ago

- 3 years, 9 months ago

Nice. Here is a similar question :)

Staff - 3 years, 9 months ago