# Start from $259^2 + 1^2 = 34 \cdot 1973$ and use the descent procedure to write the prime 1973 as a sum of two squares.

Start from $259^2 + 1^2 = 34 \cdot 1973$ and use the descent procedure to write the prime 1973 as a sum of two squares.

How to solve it using fermat descent method?

• but $2592+12=2604\ne 34\cdot 1973$ – Hagen von Eitzen Mar 10 '15 at 13:18
• – lab bhattacharjee Mar 10 '15 at 13:25

HINT:

$$259^2+1^2=34\cdot1973$$

$$34^2\cdot1973=(5^2+3^2)(259^2+1^2)$$

Use Brahmagupta–Fibonacci identity and observe that $(259\cdot5-3\cdot1,259\cdot3+5\cdot1)=34$

$34 \cdot 1973 = 259^2 + 1^2; \\ (21^2 + 1^2)(259^2 + 1^2)=(21 \cdot 259 +1)^2+(259-21)^2 \implies 13 \cdot 1973 = 7^2 + 160^2.$

$(7^2 + 4^2)(7^2 + 160^2)=(7 \cdot 7 +160 \cdot 4)^2+(160 \cdot 7-4 \cdot 7)^2 \implies 5 \cdot 1973 = 53^2 + 84^2.$

$(2^2 + 1^2)(53^2 + 84^2)=(2 \cdot 53 + 84 \cdot 1 )^2+(84 \cdot 2 -53 \cdot 1 )^2 \implies 1973 = 38^2 + 23^2.$