Math Olympiad | How to solve for "a" and "b" in this problem?

[@gaiatetuya92]

a=7のとき左辺は44を越えるので a=1,2,3,4,5,6のどれかしかない。順に与式に入れていけば a=2のときb=8, a=3のときb=5 が出る。この方が大分早い。

[@mustafasibic2954]

I have solution 1) Factor: a^2+2ab+b^2=(b+a/2)(1+2a)=44 2) Multiply by 2 (2b+a)(1+2a)=88+a 3) write b in term of a(idk this how to say it) 2b+a= (88+a)/(1+2a) 2b= (88+a)/(1+2a) - a 4b= (176+2a)/(1+2a) - 2a 4b= 1+ 175/(2a+1) - 2a 4) find b values 175=1×175=5×35=25×7 after trying we will find the solutions which is (a,b) = {(2,8),(3,5)}

[@alexanderserov3952]

(a+b) ^2 = 44+(b-1)*b Therefore 44+(b-1)*b is a full square. (b-1)*b is an even number and is a product of two consecutive numbers. Therefore b=5 and a=3 or b=8 and a=2. Sorry for my English.

[@SALogics]

Very nice trick to solve this type of problems! ❤❤

[@ericjb1030]

Case 1 could have been rejected when you found a=0 without computing b.

[@билал-ж2к]

Спасибо!!! Чудесное решение!!! ❤

[@pas6295]

a^2=9 is s a=3. Then 6b+b=44-9=35.So 7b=35. b=5. Answer ais 3 and b is 5.

[@davidshen5916]

44=A^2+2AB+B=B(2A+1)+((2A)^2-1+1)/4=(2A+1)(B+(2A-1)/4)+1/4, 44*4=(2A+1)(4B+2A-1)+1, (2A+1)(2A+4B-1)=175

[@giuseppemalaguti435]

a=-b+√(b^2-b+44)..b=-4(a=12,-4)....b=-7(a=17,-3)

[@chao541]

44 is very small. Just try 0 to 6 for a.

[@juvenalkanani2458]

Also, the couple (0; 44) verifies the original equation. Therefore it might be acceptable!

[@danikochman1351]

(a,b)=(2,8);(3,5);(12,-4);(-13,5);(-4,-4)

[@madanmohan1221]

please give answer me how to prove it = +-√-1 please🙏

[@pas6295]

a =3 b=5.

[@letsfindthejams3525]

From one equation you can not calculate the value of two ends a and b therefore all is wrong

[@pas6295]

Why do much of complications. Induction method is the easiest. Between 1 and 9 on seeing the equation by giving one value to one unknown and get the value of the other in order to satisfy the equation.

[@JunedHussain]

Positive integer is more than 1-9

[@pas6295]

@@JunedHussain The very problem is such two unknown but one equation. Since it happened to lenear one you can have multiple answers. Had it been Quadratic equation you get only their roots. But in Lenear one for two unknown to have only one answer you need one more equation.

[@JunedHussain]

@@pas6295 good

[@raj-nq8ke]

Comparison part is wrong .

[@Chapulin28]

Why?

[@yanssala2214]

Haz supuesto 3 casos pero en realidad son infinitos casos. De otro modo hablando su unica ecuacion no tiene solucion unica al tener dos incognitas. Borre esto de internet y no tupa a los estudiantes. Queridos estudiantes si a=0 entonces b = 44 si a =1 b= 4,333; si a =2 , b =8; si a=3, b =5; si a =4, b =3,111...y asi hasta infinito. El valor de a puede ser cualquier numero real y estos no son contables. No crean en este farsante y estudien matemáticas en serio.

[@sahdeolaldadsena915]

फालतू प्रश्न

[@билал-ж2к]

Спасибо!!! Чудесное решение!!! ❤