Prove that \(2005^2\) can be written as the sum of \(2\) perfect squares in at least \(4\) ways.

Prove that \(2005^2\) can be written as the sum of \(2\) perfect squares in at least \(4\) ways.

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TopNewestSolution (not mine): Notice that \(2005=|(1\pm2i)(1\pm20i)|\), so \(2005^2=|(1\pm2i)(1\pm20i)(1\pm2i)(1\pm20i)|\). Choose the \(\pm\)'s to get

\[2005^2=1037^2+1716^2=1203^2+1604^2=1995^2+200^2=1357^2+1476^2\] – Cody Johnson · 2 years, 4 months ago

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\(This\ is\ just\ one\ way.\) \(We\ are\ basically\ looking\ for\ solutions\ for\ 2005^{2}=a^{2}+b^{2}.\) \(Thus\ we\ are\ looking\ for\ a\ pythagorean\ triplet\ (a,b,2005).\) \(Now\ let\ us\ keep\ that\ aside.\) \(We\ know\ that\ (3,4,5)\ is\ a\ pythagorean\ triplet.\) \(\Longrightarrow\)\(3x,4x,5x\ is\ also\ a\ pythagorean\ triplet.\) \(now,because\ 2005=5*401,we\ take\ x\ to\ be\ 401.\) \(\Longrightarrow\)\((3*401,4*401,5*401)\ is\ also\ a\ pythagorean\ triplet.\) \(\Longrightarrow\)\(1203^{2}+1604^{2}=2005^{2}\) – Adarsh Kumar · 2 years, 4 months ago

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