\[\frac{x^5-1}{x-1} = \frac{(x-1)(x^4 +x^3 +x^2 +x +1)}{x-1} = 5^{100} +5^{75} +5^{50} +5^{25} +1 = y\] 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

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

Nice. Here is a similar question :)

Log in to reply