Find positive integers x and y such that x^2-2^y=2021.

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The video illustrates step-by-step how to find positive integers x and y such that x^2-2^y=2021.

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wrong way to do this but this is how I solved it

since 2021=43*47 then we can rewrite the whole thing as

43*47 +2^y = x^2.
Since x has to be bigger than 43 we can rewrite x as (43+n)

43*47 +2^y = (43+n)^2 => 43*47 +2^y = 43^2+43*(2n)+n^2
Dividing everything by 43 we have
47+2^y/43=43+2n+2^y/43=
4+2^y/43 = 2n+n^2/43

maybe not too obvious, but we can see that for n =2 both sides of the equation are equal with y=2. So since x=43+n then x=45 and y=2

JAzzWoods-ikvv

A guess-and-check method:
Let u = y/2
x² - 2ʸ = x² - 2²ᵘ = x² - (2ᵘ)² = (x - 2ᵘ)(x + 2ᵘ)
∴ (x - 2ᵘ)(x + 2ᵘ) = 2021 = (43)(47)
Try:
x - 2ᵘ = 43
x + 2ᵘ = 47
Adding:
2x = 90
∴ x = 45.
Subtracting:
2ᵘ⁺¹ = 4
∴ u = 1
⇒ y = 2.
45² - 2² = 2021 ✓

guyhoghton

The question seems hard to me. The method is simple, which is easy to understand.

margaretsmith

What about x=1011 and y =1010 This is also a valid answer. (1022121 - 1020100 = 2021)

chrismcgowan

Find positive integers x and y such that x^2-2^y=2021.

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