Dowód
Erche:
Udowodnić że jeżeli a
2+b
2=(a+b−c)
2 b≠c a+b≠c to
| | a−c | |
U{a2+(a−c)2}{b2+(b−c)2= |
| |
| | b−c | |
b≠c i założenia a
2+b
2=(a+b−c)
2
a
2=(a+b−c)
2−b
2=(a+b−c+b)(a+b−c−b)=(a+2b−c)(a−c)
b
2=(a+b−c)
2−a
2=(a+b−c+a)(a+b−c−a)=(2a+b−c)(b−c)
| a2+(a−c)2 | | (a+2b−c)(a−c)+(a−c)2 | |
| = |
| |
| b2+(b−c)2 | | (2a+b−c)(b−c)+(b−c)2 | |
| | a−c)(a+2b−c+a−c) | | (a−c)(2a+2b−2c | | a−c | |
= |
| = |
| = |
| |
| | (b−c)(2a+b−c+b−c) | | (b−c)(2a+2b−2c | | b−c | |
oraz a+b−c≠0