Question

find the maximum integer x such that 4^27+4^10000+4^x is a perfect square​

Answer 1

Step-by-step explanation:

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4^27 + 4^1000 +4^n

= (2)^54 + (2)^2000 + (2)^2n

= (2^27)^2 + (2^1000)^2 +2 * 2^27 * 2^1000

= (2^27)^2 + (2^1000)^2 + 2^1028

= { 2^27 + 2^1000 }^2

Hence , 2^2n = 2^1028

=> 2n = 1028

=> n= 514

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Answer 2

Step-by-step explanation:

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4^27 + 4^1000 +4^n

= (2)^54 + (2)^2000 + (2)^2n

= (2^27)^2 + (2^1000)^2 +2 * 2^27 * 2^1000

= (2^27)^2 + (2^1000)^2 + 2^1028

= { 2^27 + 2^1000 }^2

Hence , 2^2n = 2^1028

=> 2n = 1028

=> n= 514