Nice Application of Gaussian Integers

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

#NumberTheory #GaussianIntegers

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

Cody Johnson - 6 years, 9 months ago

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

Adarsh Kumar - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$