Find an example where 2^m + 2^n + 2^100 is a perfect square

[weary valley]

m not equal n not equal 100.
I first tried a few examples like 2^99+ 2^100 + 2^101, and then factoring to get 2^99(1+2+4) but then i realized that doesn't work. I also realized that i can't add different powers of 2 to get another power of (at least for 3 different powers of 2)

[deep kelp]

i mean have you considered $(2^{\frac{m}{2}} + 2^{\frac{n}{2}})^2$

[normal locust]

knief

[weary valley]

umm

i think i got an answer

lemme solve rq

i got 2^95, 2^99, and 2^100

[deep kelp]

from?

did you consider my suggestion?

[weary valley]

oih wait nvm

it doesn't work

[weary valley]

wdym by this

[deep kelp]

$(2^{\frac{m}{2}} + 2^{\frac{n}{2}})^2 = 2^m + 2 \cdot 2^{\frac{m}{2}} \cdot 2^{\frac{n}{2}} + 2^n$

[normal locust]

knief

[deep kelp]

the middle term has exponent 1 + m/2 + n/2

set that equal to 100

find suitable pairs of m and n

[weary valley]

oh ok

[deep kelp]

and im assuming we want m and n to be integers

so take m and n to be even

[weary valley]

ok

uhh where is this supposed to lead to exactly

[deep kelp]

the left side is a perfect square

therefore the right side is

[weary valley]

ohhh

[deep kelp]

so if we choose m and n such that the middle terms exponent is 100 then we've written 2^m + 2^n + 2^100 as a perfect square

[weary valley]

yea alr

[deep kelp]

and i think we want m, n, m/2, n/2 to be integers

[weary valley]

alr

[deep kelp]

so take m, n to be a pair of even integers whose sum is 198

and you're done

[weary valley]

ok thanks

[deep kelp]

no worries

[weary valley]

.close