Primality Proof

Prove or disprove that 512515251 \large \frac{5^{125}-1}{5^{25}-1} is prime.

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Note by Jihoon Kang
4 years, 6 months ago

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x51x1=(x1)(x4+x3+x2+x+1)x1=5100+575+550+525+1=y\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 3597751y not prime3597751 | y \therefore\ not \ prime You also might be able to use the pseudo prime test i.e. ay11mod(y) for gcd(a,n)=1a^{y-1} \neq 1 mod(y) \ for \ gcd(a,n) = 1

Curtis Clement - 4 years, 6 months ago

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

How did you guess that 3597751y3597751 \mid y ? Or did you use a computer?

Calvin Lin Staff - 4 years, 6 months ago

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yea I used wolfram alpha

Curtis Clement - 4 years, 6 months ago

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

Jihoon Kang - 4 years, 6 months ago

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@Jihoon Kang Great approach! Can you share your solution?

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin Certainly

Let x=525x=5^{25}

Then 512515251=x51x1=x4+x3+x2+x+1\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 x4+x3+x2+x+1=(x2+3x+1)2(5x3+10x2+5x)x^4+x^3+x^2+x+1=(x^2+3x+1)^2-(5x^3+10x^2+5x)

=(x2+3x+1)25x(x2+2x+1)=(x^2+3x+1)^2-5x(x^2+2x+1)

=(x2+3x+1)2526(x+1)2=(x^2+3x+1)^2-5^{26}(x+1)^2

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

((x2+3x+1)513(x+1))((x2+3x+1)+513(x+1))((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

Jihoon Kang - 4 years, 6 months ago

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@Jihoon Kang Cool proof. How'd you get x4+x3+x2+x+1=(x2+3x+1)2(5x3+10x2+5x)x^4+x^3+x^2+x+1=(x^2+3x+1)^2-(5x^3+10x^2+5x), though? Was it purely intuition?

Shashank Rammoorthy - 4 years, 3 months ago

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@Shashank Rammoorthy Yeah. I thought that difference of two squares may be the way to go, so I got (Ax2+Bx+1)2(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.

Jihoon Kang - 4 years, 2 months ago

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@Jihoon Kang Can you share your solution please

Curtis Clement - 4 years, 6 months ago

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@Curtis Clement I already have :)

Jihoon Kang - 4 years, 6 months ago

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Nice. Here is a similar question :)

Calvin Lin Staff - 4 years, 6 months ago

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