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Inspired by Jihoon Kang

Prove or disprove that the following number is prime:

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


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Note by Calvin Lin
2 years, 5 months ago

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Let \(a=16^{25}\), 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, 5 months ago

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@Samrat Mukhopadhyay "Sacrosanct" is such a beautiful word. Well done, too. Jake Lai · 2 years, 4 months ago

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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, 5 months ago

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@Pi Han Goh Alternatively, you could use Fermat's little theorem in Lines 1 and 2 (Line 2 is logically unnecessary, anyway): \(16^{125}=2^{500}=2^{\phi(625)}\equiv1 (\mod 625)\) . Otto Bretscher · 2 years, 5 months ago

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@Otto Bretscher 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. Pi Han Goh · 2 years, 5 months ago

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@Pi Han Goh 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... Otto Bretscher · 2 years, 5 months ago

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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, 5 months ago

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@Pi Han Goh Oh ooops. Typo. Fixed!

The question has to be much more interesting than that ... Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin I just proved it, essentially the same proof as the earlier problem :). I won't spoil it for others. But I will leave a rather interesting hint:

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, 5 months ago

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@Pi Han Goh 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 :) Jihoon Kang · 2 years, 5 months ago

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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, 5 months ago

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@Matthew Ramirez OH MY GOODNESS THAT SAYS 16. IGNORE THE PREVIOUS COMMENT. Matthew Ramirez · 2 years, 5 months ago

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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, 5 months ago

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@Pi Han Goh You wrote : x^2 + 2 x + 1 instead of x^2 + x + 1 at the 4th line Gamal Sultan · 2 years, 5 months ago

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@Gamal Sultan \( x^4 + x^3 + x^2 + x + 1 = (x^2 + x+ 1)^2 - x(x^2 + 2x + 1) \) is correct. Pi Han Goh · 2 years, 5 months ago

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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, 5 months ago

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Scratch that, it appears to check out. Matthew Ramirez · 2 years, 5 months ago

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16-1/16-1=1 Mc Tuazon · 2 years, 5 months ago

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@Mc Tuazon Note that the exponents are different, so it doesn't evaluate to 1. Calvin Lin Staff · 2 years, 5 months ago

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1 Ziad Mohamed El Doadoa · 2 years, 5 months ago

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@Ziad Mohamed El Doadoa Note that the exponents are different, so it doesn't evaluate to 1. Calvin Lin Staff · 2 years, 5 months ago

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answer is 1,048,576 Mc Tuazon · 2 years, 5 months ago

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@Mc Tuazon 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 . ) Uday Mondal · 2 years, 5 months ago

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@Uday Mondal

  1. 35÷15 is not an integer.
  2. 75÷15 is an integer and is prime.
Joel Tan · 2 years, 5 months ago

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Haha :) I guess I can prove it Jihoon Kang · 2 years, 5 months ago

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