Prove or disprove that the following number is prime:

\[ \frac{ 16 ^ {125} - 1 } { 16 ^ {25} -1 } \]

Prove or disprove that the following number is prime:

\[ \frac{ 16 ^ {125} - 1 } { 16 ^ {25} -1 } \]

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TopNewestLet \(a=16^{25}\), then the number is \(\dfrac{a^5-1}{a-1}=1+a+a^2+a^3+a^4\equiv 0(\mod{5})\ (\because a\equiv 1\mod {5})\) This shows that there is nothing sacrosanct about \(16^{25}\) anything that has a remainder \(1\) modulo 5 (in general \(n\)) will do. – Samrat Mukhopadhyay · 2 years, 3 months ago

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– Jake Lai · 2 years, 2 months ago

"Sacrosanct" is such a beautiful word. Well done, too.Log in to reply

We see that \( 16^{125} = (15 + 1)^{125} = 1 + 15 \cdot 125 + 625(\ldots ) \equiv 1 \pmod {5^4} \Rightarrow 16^{125} - 1 \equiv 0 \pmod {5^4} \)

And \(16^{25} = (15 + 1)^{25} = 1 + 25 \cdot 15 + 125 (\ldots ) \equiv 1 \pmod {5^3} \Rightarrow 16^{25} - 1 \equiv 0 \pmod {5^3} \)

But \(16^{25} = (15 + 1)^{25} = 1 + 25 \cdot 15 + 625 (\ldots ) \equiv 376 \pmod {5^4} \Rightarrow 16^{25} - 1 \not \equiv 0 \pmod {5^4} \)

So there's more powers of \(5\) that divides the numerator than the denominator, thus the number is divisible by \(5\). – Pi Han Goh · 2 years, 3 months ago

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– Otto Bretscher · 2 years, 3 months ago

Alternatively, you could use Fermat's little theorem in Lines 1 and 2 (Line 2 is logically unnecessary, anyway): \(16^{125}=2^{500}=2^{\phi(625)}\equiv1 (\mod 625)\) .Log in to reply

– Pi Han Goh · 2 years, 3 months ago

Haha, nice! I post this in a rush without proofreading it! Obviously the simplest way is to do Samrat's way. By the way, it's Euler's totient function, not Fermat's little theorem.Log in to reply

– Otto Bretscher · 2 years, 3 months ago

As a Swiss, I should be ashamed of myself to cite Fermat rather than my countryman Euler ;) I guess I was in a rush too...Log in to reply

Obviously the number is an integer because letting \(z = 25^5 \), we get

\[ \begin{eqnarray} && \frac {z^{25}-1}{z - 1} \\ &=& z^{24}+z^{23} + \ldots + z + 1 \\ &=& (z^{24} + z^{23} + z^{22} + z^{21} + z^{20} )\\ & \ & + (z^{19} + z^{18} + z^{17} + z^{16} + z^{15} ) \\ & \ & + (z^{14} + z^{13} + z^{12} + z^{11} + z^{10} ) \\ & \ & + (z^{9} + z^{8} + z^{7} + z^{6} + z^{5} ) \\ & \ & + (z^{4} + z^{3} + z^{2} + z^{1} + z^{0} ) \\ &=& (z^4 +z^3+z^2+z + 1)(1 + z^5 + z^{10} + z^{15} + z^{20} ) \\ \end{eqnarray} \]

Which is factorization of two integers, so it's not prime.

Note that with enough patience, one could show that it is divisible by \(101\) by Fermat's little theorem and exponentiation by squaring but it's a little bit tedious. – Pi Han Goh · 2 years, 3 months ago

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The question has to be much more interesting than that ... – Calvin Lin Staff · 2 years, 3 months ago

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Think about the square of a sum of the first 3 terms of a geometric progression, with the first term 1 and common ratio \(16^{25}\) – Jihoon Kang · 2 years, 3 months ago

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– Jihoon Kang · 2 years, 3 months ago

That is a really nice proof for the earlier unedited version . Thank you for sharing that. What I really like about your proof is that it also proves the edited version :)Log in to reply

If I am not mistaken, any time 6 is multiplied by itself, the resulting number's last digit will always be six. Subtracting 1 will leave 5 as the last digit, and any integer with 5 as its last digit is divisible by 5. Please correct me if I am wrong, as I only come on this site to learn. – Matthew Ramirez · 2 years, 3 months ago

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– Matthew Ramirez · 2 years, 3 months ago

OH MY GOODNESS THAT SAYS 16. IGNORE THE PREVIOUS COMMENT.Log in to reply

One could also show that \( \frac {9^{125}-1}{9^{25} - 1} \) is composite. Letting \(x = 9^{25} \)

It will simplify to this too: \(x^4 + x^3 + x^2 + x + 1 \)

Using the algebraic identity

\[ \begin{eqnarray} x^4 + x^3 + x^2 + x + 1 &=& (x^2 + x+ 1)^2 - x(x^2 + 2x + 1) \\ &=& (x^2 + x + 1)^2 - x(x+1)^2 \\ &= &(x^2 + x+ 1)^2 - 9^{25} (x+1)^2 \\ &=& (x^2 + x + 1)^2 - 3^{50} (x+1)^2 \\ &=& (x^2 + x + 1)^2 - (3^{25}(x+1))^2 \\ &=& (x^2 + x + 1 + 3^{25}(x+1)) (x^2 + x + 1 - 3^{25}(x+1)) \\ \end{eqnarray} \]

It can be easily shown that \( (x^2 + x + 1 - 3^{25}(x+1)) \ne \pm 1 \).

Because \( (x^2 + x + 1 - 3^{25}(x+1)) \) and \( (x^2 + x + 1 + 3^{25}(x+1)) \) are neither \( \ \pm 1 \), then their product is definitely not a prime number. – Pi Han Goh · 2 years, 3 months ago

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– Gamal Sultan · 2 years, 3 months ago

You wrote : x^2 + 2 x + 1 instead of x^2 + x + 1 at the 4th lineLog in to reply

– Pi Han Goh · 2 years, 3 months ago

\( x^4 + x^3 + x^2 + x + 1 = (x^2 + x+ 1)^2 - x(x^2 + 2x + 1) \) is correct.Log in to reply

the answer is 16^50+16^25+1 16 leaves a remainder of 1 when divided by 3. Therefore any power of 16 leaves a remainder 1 when divided by 3. Therefore the expression is divisible by 3 and hence is not prime. – Anand Santhanakrishnan · 2 years, 3 months ago

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Scratch that, it appears to check out. – Matthew Ramirez · 2 years, 3 months ago

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16-1/16-1=1 – Mc Tuazon · 2 years, 3 months ago

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– Calvin Lin Staff · 2 years, 3 months ago

Note that the exponents are different, so it doesn't evaluate to 1.Log in to reply

1 – Ziad Mohamed El Doadoa · 2 years, 3 months ago

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– Calvin Lin Staff · 2 years, 3 months ago

Note that the exponents are different, so it doesn't evaluate to 1.Log in to reply

answer is 1,048,576 – Mc Tuazon · 2 years, 3 months ago

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– Uday Mondal · 2 years, 3 months ago

2^1=2, 2^2=4 ,2^3=8, 2^4=16 2^5=32, therefore 2 will come again after every 4 times, now (16)^125=2^500,(dividing 500 by 4 means remainder will zero imply the unit digit of 2^500 will gives 6, now if we subtractions 1 from 6 we will get 5 and similar way id case of denominator we will get the unit digit 5 . now as the numerator is > denominator with unit digit 5 in both the cases . therefore numerator will divisible by denominator which imply the fraction will not be a prime number it will be a composite number . )Log in to reply

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Haha :) I guess I can prove it – Jihoon Kang · 2 years, 3 months ago

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