## Primes and squares

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• Mark wrote: 100521=3^4*17*73 For a prime to be sum of 2 squares ,one must
Message 1 of 1 , Apr 4 11:24 AM
Mark wrote:
<<Compare this to the composite number 100521.

x^2 + y^2 = 100521 has 2 solutions.>>
100521=3^4*17*73

For a prime to be sum of 2 squares ,one must be odd and the other even,else the prime would be an even number.

(2*k+1)^2+(2*k1)^2=4k2+1

So, a prime of the form 4*k+3 can never be sum of 2 squares.

Now ,the Fibonacci very simple identity:

(a^2+b^2)*(c^2+d^2)=(a*c+/-b*d)^2+(a*d-/+b*c)^2

shows that a product of two different 4*k+1 primes is always a sum of two squares in two ways----->1241=17*73=(1^2+4^2)*(3^2+8^2)=35^2+4^2=29^2+20^2

and 100251=3^4*1241=36^2+315^2=261^2+180^2

If (a^2+b^2) and (c^2+d^2) were not 4k+1 primes,they would form more sums of 2 squares in at lest 2 ways,until we reach a prime.

Else there would not be 4*k+1 primes at all.

And so Fermat alone; no need of complex numbers, no need of Gauss or Euler did prove this theorem.

Now, if you solve completely the equation a^2+b^2=c^2+d^2 (with a free parameter, say: a which can take any value). And i solved it, then you can find all the 4*k+1 primes without any sieving.

Though time and/or space required might be wider than traditional seiving methods. (my knoweledge is too short to be sure of that).

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