Question

how to solve x^2+y^2=208(x-y) such x,y are natural numbers

Answer 1

Step-by-step explanation:

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x²-y²=208(x-y)

x²-y²=208x-208y

x²-y²-208x+208y=0

x²-y²-208(x-y)=0---> equation 1

If y=0

x²-208(x-0)=0

x²-208x=0

x²=208x

x=208x/x

x=208

If x=0

-y²+208y=0

-y²=-208y

y²=208y

y=208y/y

y=208

Thank you

@ Twilight Astro ✌️☺️♥️

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Answer 2

Step-by-step explanation:

GIVEN THE TASK,

how to solve x^2+y^2=208(x-y) such x,y are natural numbers,

According to the question

x²-y²=208(x-y)

x²-y²=208x-208y

transferred the equation

x²-y²-208x+208y=0

x²-y²-208x+208y=0x²-y²-208(x-y)=0..........(1 )