My Brother asked me what is the smallest prime number of the form $A^{B^C}+D^{E^F}$ where A,B,C are three distinct prime numbers, and D,E,F are 3 distinct primes that is Permutations of those 3 primes.I realize that we must arrange the exponents to be as smallest as possible in order to get smallest prime as possible.So what is the smallest such prime? (and second smallest :-)
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1Note that large primes are odd, so $A$ or $D$ must be equal to $2$ – Mark Bennet Jun 15 '15 at 07:48
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@MarkBennet ,yes,but I am not sure that F must be 2 – Kenan D Jun 15 '15 at 07:49
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2Curiously, $2^{(5^3)}+3^{(2^5)}$ is prime. – bonsoon Jun 15 '15 at 08:03
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@bonsoon are you sure it is a prime ? – Kenan D Jun 15 '15 at 08:04
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At least according to WolframAlpha: http://www.wolframalpha.com/input/?i=isprime%282%5E5%5E3%2B3%5E2%5E5%29 – bonsoon Jun 15 '15 at 08:05
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1@NeilRoy 1 isn't prime. – Jimmy360 Jun 15 '15 at 08:26
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@NeilRoy A, B, C, D, E, and F ALL must be prime. – Jimmy360 Jun 16 '15 at 01:32
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$A = 2$ OR $D = 2$
$A \neq D$, $B \neq E$, $C \neq F$
$a^n$ generally increases faster as $n \rightarrow \infty$ than as $a$ does.
If we say that $A = 2$, then $A^{B^C}$ must be even. Therefore, $D^{E^F}$ must be odd.
Therefore we have $2^{B^C} + D^{2^F}$ OR $2^{B^C} + D^{F^2}$
Let's try $2^{5^3}+3^{2^5}$. It is prime! Are there any smaller primes of the specified form? Taking $2^{B^C} + D^{2^F}$, we have two cases: is $5^3 < 3^5$ and is $2^5 < 5^2$. The first cases is true, but the second case is not. Therefore, let's try varying the second case. Is $2^{5^3}+3^{5^2}$ prime? It is not! Therefore, $2^{5^3}+3^{2^5}$ is the answer!
Jimmy360
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If $D$ is odd, every power of $D$ is odd, regardless of whether the exponent is odd or even. – Joffan Jun 15 '15 at 18:51
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If we assume the first term is $2^{X^Y}$, the second term can be either $X^{Y^2}$ or $Y^{2^X}$ – Joffan Jun 16 '15 at 02:10
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