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BIG square...

Wilmer

In Memoriam
Mar 19, 2012
376
Working 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.
 

Amer

Active member
Mar 1, 2012
275
I do not know if this work for you


to make sure it is an integer I took the floor of the square root here
check this
and this

to make sure it is an integer I took the floor of the square root here
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Working 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.
For the manageable example
$$4^3 + 4^7 + 4^{10} = 4^3(1 + 4^4 + 4^7) = 2^6(1 + 2^8 + 2^{14}) = \bigl(2^3(1+2^7)\bigr)^2.$$
Now try the same procedure with your bigger numbers.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks loads, Amer.
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.
 

Wilmer

In Memoriam
Mar 19, 2012
376
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
 
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