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Prime or composite?

Is the number 258+15\displaystyle \frac{2^{58} + 1}{5} prime or composite?

[Ripped from the Oxford Maths Institute puzzle of the day, so one would assume 5 is a factor of the numerator. I am genuinely interested in how others might go about solving this... I am having some trouble.]

EDIT: For completeness http://www.maths.ox.ac.uk/noticeboard/last_puzzle
You know you're a virgin when you attempt the Oxford Mathematics puzzle of the day...
Reply 2
UCASquestions
You know you're a virgin when you attempt the Oxford Mathematics puzzle of the day...


Well, I know of it because I work at the Oxford Centre For Nonlinear PDE (and I got laid [twice] this morning, not that it matters :p:).
Dream Eater
Well, I know of it because I work at the Oxford Centre For Nonlinear PDE (and I got laid [twice] this morning, not that it matters :p:).


:rofl:, too much info. :rofl2:
Dream Eater
Is the number 258+15\displaystyle \frac{2^{58} + 1}{5} prime or composite?

[Ripped from the Oxford Maths Institute puzzle of the day, so one would assume 5 is a factor of the numerator. I am genuinely interested in how others might go about solving this... I am having some trouble.]

EDIT: For completeness http://www.maths.ox.ac.uk/noticeboard/last_puzzle


My first thoughts are:

Spoiler


Spoiler


Spoiler



Not sure if this is going anywhere! :smile:
And yeah too much info there :smile:
composite.

mini hint

Spoiler



massive hint

Spoiler

Reply 6
Dream Eater
Is the number 258+15\displaystyle \frac{2^{58} + 1}{5} prime or composite?

[Ripped from the Oxford Maths Institute puzzle of the day, so one would assume 5 is a factor of the numerator. I am genuinely interested in how others might go about solving this... I am having some trouble.]

EDIT: For completeness http://www.maths.ox.ac.uk/noticeboard/last_puzzle


258+122+10(mod5)2^{58} + 1 \equiv 2^2 + 1 \equiv 0 \pmod 5, so the numerator is indeed divisible by 5. Alternatively, note that x58+1=(x2+1)(x56x54++1)x^{58} + 1 = (x^2 + 1)(x^{56} - x^{54} + \cdots + 1).

Beyond that, I have no idea.
Dream Eater
Well, I know of it because I work at the Oxford Centre For Nonlinear PDE (and I got laid [twice] this morning, not that it matters :p:).

was it good sex though? that's what we all want to know.
Totally Tom
was it good sex though? that's what we all want to know.


:rofl:, aahhaahahahha. :rofl3:
Reply 9
Totally Tom
massive hint
258+1=(229+1)22302^{58}+1=(2^{29}+1)^2-2^{30} now think difference of 2 squares...


Yay! I thought it involved Fermat's Factorisation method but I couldn't think of where to apply it. Cheers :smile:
Reply 10
Totally Tom
was it good sex though? that's what we all want to know.


Bloody fantastic. Had tea afterwards and everything.
Dream Eater
Bloody fantastic. Had tea afterwards and everything.

:hi5:
Dream Eater
Yay! I thought it involved Fermat's Factorisation method but I couldn't think of where to apply it. Cheers :smile:

try this one:

is 376+45\displaystyle\frac{3^{76}+4}{5} prime or composite?

using a different method to before (unfortunately the other method works here too).
Reply 13
Totally Tom
try this one:

is \displaystyle\frac{3^{76}+4}{5} prime or composite?

using a different method to before (unfortunately the other method works here too).


Ah yes, I can do this. This is actually how I tried to initially tackle the first one. Using modulo arithmetic one can check that:

313mod253^1 \: \equiv \: 3 \mod 25
329mod253^2 \: \equiv \: 9 \mod 25
332mod253^3 \: \equiv \: 2 \mod 25

......

3201mod253^{20} \: \equiv \: 1 \mod 25
3213mod253^{21} \: \equiv \: 3 \mod 25

with 3^21 being the first 'cycle' back to 3 mod 25. So this cycle repeats itself every 20 powers. Hence, using the fact that 76 = 16 mod 20:

376316mod2521mod253^{76} \: \equiv \: 3^{16} \mod 25 \: \equiv \: 21 \mod 25

But 21 mod 25 + 4 = 0 mod 25. Hence 376+43^{76} + 4 is divisible by 25. Hence the original expression is divisible by a further 5. Thus it is not prime.

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