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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 $d$?

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up vote 7 down vote accepted

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

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I already know all this. I need to know how I am supposed to break up $19^{10}$ into a sum of two squares. For example: $n=19^{10}=(c^2+d^2)$. – maroon.elephants Nov 6 '12 at 21:35
@maroon.elephants: you don't need to break $19^{10}$ into the sum of two squares. Given that you can break $13$, just multiply $4$ and $9$ by (what?) – Ross Millikan Nov 6 '12 at 21:37
@maroon.elephants Take a second to read Mark's answer carefully. He's given you the answer. Remember that the product of two squares is again a square. – EuYu Nov 6 '12 at 21:37
Ohh, that makes so much sense now. $19^{10}*(3^2+2^2)=19^{10}*13$. – maroon.elephants Nov 6 '12 at 21:41

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
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