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Clever approach! Thanks for sharing!

Great. Finding rational solutions is converted to finding integer solutions.

There are no positive solutions.

a=b=c=1, there may be others.

Easy

Nice video. Thank you for sharing

Great

Great. Thank you for sharing

Nice video

Thanks for sharing.

Great

x^4 - x + 1/2 = 1/4 (4x^4-4x^2+1+4x^2-4x+1) = 1/4 (2x^2-1)^2 + 1/4 (2x -1)^2 >= 0 And the expression is only zero when both squares are simultaneously zero (never)

Please.

Subtract 4^x = (2^x)^2 from both sides of the equation and factor the difference of squares.

Great. Thank you!

Nice method

Great

nice video

Great

Thanks for the detailed explanation of the solution !

Great

Great

Great. Thank you

Nice method

While this is cool, is there a more general approach?

Great

Nice solution! Thanks

Nice video.

Great

Thanks for sharing!

Great

Great

Nice video

Wow, that was hard. Finding one solution was easy. The larger number is odd while the smaller number is even and a bit larger than half the larger, so consider the list of Pythagorean triples to find 15,8,17. Thus n=7 is one of the solutions.

The examples suggest that the choice of sign in the numerator of n_k should be -(-1)^k. It would be a nice result to show that this is the case generally.

Great

Without solving quadratic equation for root, you can directly reduce it to x^2-d y^2=1

Nice proof! Thanks

Pell equation! Very good explanation! Thank you for sharing!

Great

n=4

Great! Nice video

GOOD ❤❤❤

x=0, y=-5

By observation,x,y are positive odd numbers. Otherwise 2022=2025-3=2025-4+1 =45^2-2^2+1=1+47×43=1+(43+4)43 =1+43×43+4×43=1^2+43^2+4×43. (x,y)=(1,43) or (43,1) for positive integers.

x^2+4xy+y^2-2022=0 x=-2y(+/-)√(3y^2+2022) Since x & y is an integer so discriminant is a perfect square Take 3y^2+2022=z^2 z^2-3y^2=2022(Exercise for u )

Great

These are the only solutions ?

Nice solution! Thanks.

If x = 6 then 216 - 30y + y^3 = 2023 so y^3 -30y -1807 = 0 so y^3 - 13y^2 + 13y^2 - 169y + 139y -1807 = 0 but 139 times 13 is 1807 so (y-13)(y^2 + 13y +139)=0 and since the original equation was symmetric in x and y we have x=6 and y =13 or x=13 and y=6.