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

@Shashank Rammoorthy
–
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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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?

Log in to reply

yea I used wolfram alpha

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Nice. Here is a similar question :)

Log in to reply