1996 IMO Problem #4

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  1. Mr. Math
  2. 1996 IMO
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The much easier way would be just solve for a and b and get 481b=16x^2-15y^2 and 481a = 15x^2+16y^2 and therefore 481|16x^2-15y^2 and 481|15x^2+16y^2. Some simple modular arithmetic we can conclude 481|x^2 and 481|y^2.

cr
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Hey Mr. Nal! Love your lectures, they are quite helpful and fun! Could you please do some modern olympiad problems? (ISL, etc.) Thank you!

realquarterb
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Your videos are great... Good problems and nice solutions... Thanks a lot

khaledqaraman
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How does one think of considering u^4 + v^4? It seems very arbitrary to come up with. In the process of problem solving, how does one get to the ‘crux’ as you called it?

JohnSmith-vqho
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Just when I thought I was good at Maths

eroskuikel
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When I was in 8th grade my teacher gave me this problem, I found that Min is 481^2 but I didn't proof it:]
I didn't know that it is IMO problem

mardymsyz