# Nice Application of Gaussian Integers

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

Note by Cody Johnson
3 years, 10 months ago

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Solution (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

$2005^2=1037^2+1716^2=1203^2+1604^2=1995^2+200^2=1357^2+1476^2$

- 3 years, 10 months ago

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

- 3 years, 10 months ago