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Write the following numbers as a sum of two squares:

$(a) 13\times17$ $13\times17=221$ $m=13=(3^2+2^2) \space \text{and} \space n=17=(4^2+1^2)$ $[(3)(4)+(2)(1)]^2+[(3)(1)-(4)(2)]^2=14^2+(-5)^2=221$

I had no problem solving this one. However, the next problem is giving me trouble because $19^{10}$ is such a large number.

$(b) 13\times19^{10}$

$13\times19^{10}=79703861351413$ I can only find this number using wolfram. My calculator wont display all the digits. Which is a problem.

$m=13=(3^2+2^2) \space \text{and} \space n=19^{10}=(c^2+d^2)$

Is there a easy / simple method I can use to find $c$ and $d$?

2 Answers 2

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Hint: 13=9+4, $19^{10} $ is a square.

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    Ohh, that makes so much sense now. $19^{10}*(3^2+2^2)=19^{10}*13$.2012-11-06
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Since the excellent answer by Mark Bennet has been accepted I'll undelete my stupid one:

write $19$ as a sum of two squares then

  • write $10$ in binary 1010 so $19^{10} = (19^2)(((19^2)^2)^2$
  • use the identity $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$ repeatedly