TMMATE problema 3

Moderators: Bogdan Posa, Laurian Filip

Post Reply
User avatar
Beniamin Bogosel
Co-admin
Posts: 710
Joined: Fri Mar 07, 2008 12:01 am
Location: Timisoara sau Sofronea (Arad)
Contact:
Contact Beniamin Bogosel

TMMATE problema 3

Post by Beniamin Bogosel »

Gasiti toate numerele intregi \( a,b \) pentru care \( 13a^2+10ab+4b^2=441 \).

TMMATE 2009
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present. :)

Blog
Top
Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Se observa ca daca (a,b) e solutie atunci (-a,-b) este solutie.

Asadar sa gasim mai intai solutiile cu a natural.

Deasemenea a trebuie sa fie neaparat impar si scriind egalitatea sub forma

\( (4b+5a)^2+27a^2=1764 \)

deducem ca \( a\in\{1,3,5,7\} \) apoi gasim b , s.a.m.d.....
Top
Post Reply

Return to “Clasa a VIII-a”