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

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

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

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

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TopNewestHint :- \(x^4 + 4y^4 = (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy) \) – Siddhartha Srivastava · 1 year, 9 months ago

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– Calvin Lin Staff · 1 year, 9 months ago

Sophie Germain to the rescue!Log in to reply

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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 · 1 year, 9 months ago

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– Vincent Miller Moral · 1 year, 9 months ago

I'm a newbie in Number Theory. Thanks.Log in to reply