This is a sequel to my earlier post - Numbers: Not everything is magic.

If I show the numbers 1024, 2048, 4096... to a software geek, the immediate answer would be around bits and bytes. But these numbers have another peculiar pattern.

1024 is the first positive integer to satisfy the condition.

a

^{n+2}+ a^{n}= (2a)^{n}It is followed by the remaining numbers. Check the following:

1024

^{22}+ 1024^{20}= 2048^{20}^{}2048

^{24}+ 2048

^{22}= 4096

^{22}

^{}4096

^{26}+ 4096

^{24}= 8192

^{24}

^{}8192

^{28}+ 8192

^{26}= 16384

^{26}

^{}16384

^{30}+ 16384

^{28}= 32768

^{28}

^{}……

Now the hack to find it.

a

^{n+2}+ a^{n}= (2a)^{n}^{}=>a

^{n}(a

^{2}+ 1) = 2

^{ n}a

^{n}

^{}=>a

^{2}+ 1 = 2

^{ n}

^{}=>a= (2

^{ n }- 1)

^{1/2}

Now, let us find out a way to solve the below equation:

a

^{n+x}+ a^{n}= (2a)^{n}^{}=>a

^{n}(a

^{x}+ 1) = 2

^{ n}a

^{n}

^{}=>a

^{x}+ 1 = 2

^{ n}

^{}=>a= (2

^{ n }- 1)

^{1/x}

So, for n and n+4, the first positive integer to satisfy is 64.

Last but not the least for the post.

a

^{n+x}+ a^{n}= (k*a)^{n}^{}=>a

^{n}(a

^{x}+ 1) = k

^{ n}a

^{n}

^{}=>a

^{x}+ 1 = k

^{ n}

^{}=>a= (k

^{ n }- 1)

^{1/x}

So for k=3 and x = 1, the numbers are 2, 8, 26, 80...

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