# 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
3 years ago

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

Sophie Germain to the rescue!

Staff - 3 years ago

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 3^4 = 1 (mod 10) 3^4^2 = 1 (mod 10) . . . 3^4^5 = 1 (mod 10) 4^5 = 4 (mod 10) 4^5^2 = 4^5 (mod 10) = 4 (mod 10) . . . 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 

- 3 years ago

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.

Staff - 3 years ago

I'm a newbie in Number Theory. Thanks.

- 3 years ago