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Interesting Problem!

Prove that : \( 3^{4^{5}} + 4^{5^{6}} = a.b \)

Which a,b is positive integer and each have atleast 2015 digits

Note by Rony Phong
2 years, 4 months ago

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Hint :- \(x^4 + 4y^4 = (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy) \)

Siddhartha Srivastava - 2 years, 4 months ago

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Sophie Germain to the rescue!

Calvin Lin Staff - 2 years, 4 months ago

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3^4 = 1 (mod 10)
3^4^2 = 1 (mod 10)
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3^4^5 = 1 (mod 10)

4^5 = 4 (mod 10)
4^5^2 = 4^5 (mod 10) = 4 (mod 10)
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.
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4^5^6  = 4 (mod 10)

3^4^5 + 4^5^6 = (1+4) (mod 10) = 5 (mod 10)

Last digit is 5 so 3^4^5 + 4^5^6 is not prime

Vincent Miller Moral - 2 years, 4 months ago

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Be careful, that isn't the way that tower of exponents work. \( 3^ { 4 ^ 2 } = 3 ^ {16 } = \left( 3^4 \right) ^ 4 \). However, the series of modular arithmetic statements that you wrote are still true.

Note that you are supposed to show that they are the product of 2 digits with > 2015 digits. You have shown that it is a multiple of 5, but that is not sufficient to show that \(a, b\) exist.

Calvin Lin Staff - 2 years, 4 months ago

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I'm a newbie in Number Theory. Thanks.

Vincent Miller Moral - 2 years, 4 months ago

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