Hello,please tell me how to proceed on this,
Find all natural numbers x,y,z such that
4^x + 4^y + 4^z is a perfect square.
Archishman.

(x) Find natural x, y ³ 1 with 4^x+4^y+1 is a perfect square.

Can you see how to answer (x)?

Kerwin

Hi Kerwin,
The defn. of natural numbers as I have always known(in all text books that i have,read)is that it includes all +ve integers greater than 0.
Again,if we divide by 4 throughout,how can we get a square?
Thanks for your time.
Archishman.

well,
the equation (x) above can be written as .....

(2^x )² + 2^2y + 1

if i write 2y=x+1 then the equation turns out as...
(2^x )² + 2x2^x x1 + 1
i.e(2^x +1)² ...a perfect square.....

therefore,for all x,y such that 2y=x+1 we see that the number is a perfect square...

am i right Kerwin??
love arun

That's right, but since we have to find all solutions to the original problem, we must find all solutions (and prove it).

Kerwin

Hi Kerwin,
Please do the problem with the definition that natural numbers the numbers 1,2,3,...,n,...
Next,could you explain what you meant by
"if x,y,z are all> 0 then we can divide by 4 and still get a square."
Thanks,
Archie.

If x,y,z are all > 0, we can conclude that 4^x +4^y +4^z is even and is a square, so we can divide by 4 and still get a square.

The bit about whether to include zero or not is not that important. Since we showed if and only if (x,y,z) is a solution, then (x+1,y+1,z+1) is a solution, you just have to be careful when giving your final answer.

BTW, we still have not proved those are the only solution....

Hope this clear up

Kerwin

well,
i would it a bit a clearer....for you archishman..

let 4^x +4^y +4^z be a square....
let us take 4 common here....then....we get...
4x[4^x-1 +4^y-1 +4^z-1 ]

now 4 is a square..therefore...[4^x-1 +4^y-1 +4^z-1 ] must be a square if [4^x +4^y +4^z ] is to be a square....

like this we can continuously divide by 4 and still get many square....this is what is Kerwin is saying..(am i right Kerwin??)

hope that's clear
love arun

Yes, that's right. Also, your solution set is essentially complete. The complete set of non-negative integers (x,y,z) for which x ³ y ³ z and 4^x+4^y+4^z is a square are parametrised by (2a+b+1,a+b+1,b) where a, b are non-negative integers. (So if you want a solution in naturals, you just need to take b positive.)

As above, we can set z=0, WLOG. It is easy to check x, y are positive, and it then suffices to prove that 2y=x+1. Now since 4^x+4^y+1 is a square which is larger than (2^x)² we can write:

4^x+4^y+1=(2^x+a)²

for some positive integer a.

Expanding and dividing through by 2^x+1 we obtain

2^2y-x-1 = a + (a² - 1)2^-x-1.

The right hand side is at least 1. It follows that the left hand side is an integer.

Therefore (a² - 1)2^-x-1 is integral. Note that a-1 and a+1 can't both be multiples of 4, so either 2^x|a-1 or 2^x|a+1.

Therefore unless a = 1, we must have a ³ 2^x-1 in which case we find:

2^2x+1 ³ 4^x + 4^y = (2^x + a)² - 1 ³ (2^x+1-1)² - 1 = 2×2^2x+1 - 2^x+2

This gives 2^x+2 ³ 2^2x+1, 1 ³ x so x = y = 1, therefore 4 + 4 + 1 = (2 + a)² which still gives a = 1.

So all solutions must have a = 1, so 2^2y-x-1 = 1, and 2y = x+1 so we're done.

I think Kerwin has a proof by writing the number in base 4; that may be a bit more intuitive.