Japanese | Can you solve this? | Nice Math Olympiad Algebra Problem

[@bogdangarkusha8727]

a=b^2 - 13
a=(b+√13)(b-√13)
(b+√13)^2(b-√13)^2 = b+13
(b^2+2b√13+13)(b^2-2b√13+13) = b+13
let u=b^2+13, v=2b√13
(u+v)(u-v)=b+13
u^2-v^2=b+13
reverting
(b^2+13)^2 - 42b^2 = b+13
(b^2+13)^2 = 42b^2 + b + 13
b^4 + 26b^2 + 169 = 42b^2 + b + 13
b^4 - 16b^2 - b + 156 = 0
to solve, we can use a ready formula for quartic equation
and something seems off, because both vid solution and this form should be true, whereas using wolfram alpha
b_1,2=-3.2008+-1.5242i
b_3,4=3.2008+-1.4721i
(vid solution is true, of course 😊, (-4)^2 = 13+3)

[@rabotaakk-nw9nm]

2:15 a+b+1=0; b=-a-1 -> (1)
a²=(-a-1)+13; a²+a-12=0 => a1=-4; a2=3
b1=-(-4)-1=3; b2=-3-1=-4
(a, b)=(-4, 3),( 3,-4) 😁

[@johnscovill4783]

Hi, Master T. When I graph this on Desmos, I see your solutions. And two additional ones.

[@MasterTMathsClass]

@@johnscovill4783 Nice 🎉

[@prollysine]

b=a^2-13 , a^4 +/- a^3 -26a^2 - a + 156 = 0 , (a+4)(a^3-4a^2-10a+39)=0 , a= -4 , a^3-4a^2-10a+39=0 , (a-3)(a^2-a-13)=0 , a= 3 ,
+1 +4 +1 -3
-4 -16 -1 +3
-10 -40 -13 +39=0 ,
+39 +156 = 0 , a^2-a-13=0 , a=(1+/-V(1+52))=2 , a=(1+/-V53)/2 , /not integer/not a solu ,
b=a^2-13 , solu , (a, b) = (-4 , 3 ) , ( 3 , -4) ,
test , (a, b) = (-4 , 3 ) , (-4)^2=b+13 , 16=3+13 , -> 16 , OK , test , (a, b) = ( 3 , -4) , 3^2= -4+13 , -> 9 , OK ,

[@MasterTMathsClass]

Nice