Question

determinati a,b naturale stiind 20a²+16b²=2016

Answer 1

20a²+16b²=2016
5a²+4b²=504
5a²+4b²=500+4
5a²+4b²=5×10²+4×1²
a=10
b=1

Answer 2

5a²+4b²=504
5a²+4b²=5*100 +4*1
 se observa ca a²=100 si b²=1 verifica eciatia de gradul 2 cu 2 necunoscute

 
a²=100⇒a=+/√100=+-10 dar a∈n, deci a=10
b²=1⇒b=+-√1=+-1 dar b∈N, deci b=1
 

Ecuatia are o infinitatede solutii in (R+)x( R*)
trebuie sa verificam daca are  in NxN, siu alte solutiidecat cea vizibila a=10, b=1

a=√((504-4b²)/5)

pt b=2, 3,4,5, 6,7,8,9,10,11,vom obtine, respectiv
a=√488/5∉N
a=√468/5∉N
a=√440/5=√88∉N
a=√404/5∉N
a=√(360/5)=√72∉N
a=√308/5∉N
a=√248/5∉N
a=√180/5=√36=6∈N deci b=9, a =6 este solutie buna
Verificare: 20*36+16*81=2016 adevarat

a=√104/5∉N
a=√20/5=√4=2∈N, deci b=11, a=2 este solutie buna
verificare ;20*4 +16*121=2016 adevarat


ptb≥12 nu mai berificam, deaoirece 504-4b²<0, a∉R

asadar (a;b) ∈ {(10;1);(6;9);(2;11)}