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- Thread starter Wilmer
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- Feb 7, 2012

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For the manageable exampleWorking with a problem with huge integers;

is 4^27 + 4^500 + 4^972 a square? (length of that is 586 digits!)

Manageable(!) example:

4^3 + 4^7 + 4^10 = 1032^2

Thanks for any help; couldn't find an online calculator that can handle that.

$$4^3 + 4^7 + 4^{10} = 4^3(1 + 4^4 + 4^7) = 2^6(1 + 2^8 + 2^{14}) = \bigl(2^3(1+2^7)\bigr)^2.$$

Now try the same procedure with your bigger numbers.

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I'll make SURE to remember Wolfram!

This was the original problem:

4^27 + 4^500 + 4^n = k^2

where n and k are both positive integers.

What is n?

This was the given solution:

4^27 + 4^500 + 4^n

= 4^27 (1 + 2^946 + 4^(n - 27))

= 4^27 (1 + 2.2^945 + (2^(n-27))^2)

= 4^27 (1 + 2.2^945 + (2^945)^2) ***********

= ((2^27)(1 + 2^945))^2

= k^2 where k = (2^27)(1 + 2^945)

so it must indicate n - 27 = 945

it means n = 27 + 945 = 972

I can follow almost all of it; except the switch from n-27 to 945 ***********

Can you see why? Thanks.

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Thanks Opal; now quite clear. Slightly differently:

4^27 + 4^500 + 4^n

= 2^54 + 2^1000 + 2^(2n)

= 2^54[1 + 2^946 + 2^(2n - 54)] : A

= 2^54[1 + 2^(n - 27)]^2

= 2^54[1 + 2^(n - 26) + 2^(2n - 54)] : B

A, B: 2^54[1 + 2^946 + 2^(2n - 54)] = 2^54[1 + 2^(n - 26) + 2^(2n - 54)]

So:

2^946 = 2^(n - 26)

n - 26 = 946

n = 972

In other words, using 4^u + 4^v + 4^n :

n = 2v - u - 1

4^27 + 4^500 + 4^n

= 2^54 + 2^1000 + 2^(2n)

= 2^54[1 + 2^946 + 2^(2n - 54)] : A

= 2^54[1 + 2^(n - 27)]^2

= 2^54[1 + 2^(n - 26) + 2^(2n - 54)] : B

A, B: 2^54[1 + 2^946 + 2^(2n - 54)] = 2^54[1 + 2^(n - 26) + 2^(2n - 54)]

So:

2^946 = 2^(n - 26)

n - 26 = 946

n = 972

In other words, using 4^u + 4^v + 4^n :

n = 2v - u - 1

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