Math Olympiad | A Nice Algebra Problem | How to solve for 'a' and 'b' in this problem?

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[@user-pd7js7cy9m]

Thank You. But after 3:08 it’s possible a little differently. (1) a^2-b=111 ; (2) b^2-a=111 ; (3) a=-b-1 . We substitute (3) in (2) . We get : (4) b^2+b-10*11=0 . Therefore : (5) b1=10 , b2=-11 . ( thanks to Viett ) . We substitute (5) in (3) - we get Your answer.
With respect, Lidiy

[@BN-hy1nd]

I think that there is no need to add eqn1 to eqn 2 . Just use your a+b+1=0 which means that a = -b -1 then substitute a = -b -1 into eqn1 you will get
a = (-b-1) substitute this into eqn 1 and you get
(-b-1)^2 -b = 111 expanding you will get
(b^2 + 2b +1) -b = 111 simplifying gives
b^2+b+1 = 111 => b^2+b+1 -111 =0 which then gives
b^2+b -110 =0 => b^2+11b-10b-110 =0 => b(b+11)-10(b+11)
(b-10)(b+11) =0. so b has two values (10 and - 11)
substitute these values in eqn 2 to get a with two values of -11 and +10
so (a,b) =(-11,10) and (a,b) = (10, -11). Hope this helps

[@guydorian1828]

I stopped at equation 3. Very brilliant!

[@romeusricardo3359]

Good job

[@learncommunolizer]

Thank you very much!!!

[@zvonimirkujundzic6867]

Beautiful!🎉

[@learncommunolizer]

Thank you very much!!!

[@jeffreykalb9752]

Or just make a simple substitution to get a quartic equation, use the possible rational zeros theorem to identify possible roots, and do a little synthetic division...

[@zdrastvutye]

how about bbc basic?
10 sw=1:a=sw:goto 40
20 b=a*a-111:dgu1=b^2/a^2:dgu2=1/a:dgu3=111/a^2:dg=dgu1-dgu2-dgu3:return
40 gosub 20
50 dg1=dg:a1=a:a=a+sw:a2=a:gosub 20:if dg1*dg>0 then 50
print a,b

[@alexanderbrinkley4332]

This is the fourth time you have solved the same problem, just with different values for the right hand side: 111, 133, 41, 33.
You are taking longer each time.

[@ilm_izla166]

❤

[@learncommunolizer]

❤️❤️❤️

[@user-ec5ip3vp2r]

(-11;10)(10;-11)

[@cliffordabrahamonyedikachi8175]

A = 24642.
B = 12321.