Observe that 244=3^5+1, and the solution is obvious: (x,y)=(4,9) and by symmetry of the problem (x,y)=(9,4). Or: s+x-4 and t:=y-4, t=5-s from the second equation. Then the first equation, after multiplication by 3^s leads to a quadratic equation in 3^s: (3^s)^2-244*3^s+244=0, which leads to 3^s=[244+-sqrt(244^2-4*243)]/2. But 244^2-4*243=244^2-4*244+4, and so 3^s=0 or 3^s=5 ==> (x,y)=(4,9) or (x,y)=(9,4).
3ˣ⁻⁴ = a 3ʸ⁻⁴ = b a + b = 244 ab = 3⁵ = 243 t² - 244t + 243 = 0 (a, b are the roots) (t - 1)(t - 243) = 0 (a, b) = {(1, 243); (243, 1) a = 1 => 3ˣ⁻⁴ = 1 => x = 4 b = 243 => 3ʸ⁻⁴ = 243 => y = 9 a = 243 => x = 9 b = 1 => y = 4 *(x, y) = {(4, 9); (9, 4)}*
OK, j'ai trouvé les deux solutions (x,y)=(4,9) et (9,4). Même technique en procédant à deux changements de variables. La première, x-4=a, y-4=b, puis multiplication de tous les termes pas 3^a, mise en facteur pour obtenir une équation du second degré, et changement de variable à nouveau 3^a = A (A par exemple) et le tour est joué.
Observe that 244=3^5+1, and the solution is obvious: (x,y)=(4,9) and by symmetry of the problem (x,y)=(9,4). Or: s+x-4 and t:=y-4, t=5-s from the second equation. Then the first equation, after multiplication by 3^s leads to a quadratic equation in 3^s:
(3^s)^2-244*3^s+244=0, which leads to 3^s=[244+-sqrt(244^2-4*243)]/2.
But 244^2-4*243=244^2-4*244+4, and so 3^s=0 or 3^s=5 ==> (x,y)=(4,9) or (x,y)=(9,4).
3ˣ⁻⁴ = a
3ʸ⁻⁴ = b
a + b = 244
ab = 3⁵ = 243
t² - 244t + 243 = 0 (a, b are the roots)
(t - 1)(t - 243) = 0
(a, b) = {(1, 243); (243, 1)
a = 1 => 3ˣ⁻⁴ = 1 => x = 4
b = 243 => 3ʸ⁻⁴ = 243 => y = 9
a = 243 => x = 9
b = 1 => y = 4
*(x, y) = {(4, 9); (9, 4)}*
244=3^5+1 --> 243+1=3^5+3^0 , , case 1 , x-4=5 , x=9 , & y-4=0 , y=4 , case 2 , x-4=0 , x=4 , & y-4=5 , y=9 , 9+4=13 ,
solu x , y : 9 , 4 or 4 , 9 , OK ,
The answer is (x,y)=(4,9). And man this just another nice probelm and trick that I better start practicing on!!!
Thanks 👍💯. You're welcome 😊🙂
3^(x - 4) + 3^(y - 4) = 244 = 3^5 + 1
x + y = 13
or
3^x + 3^y = 3^9 + 3^4 = 3^p + 3^q (p = 9, q = 4 => p + q = 13)
x + y = 13
Therefore
(x,y) : (9, 4) and (4, 9) are solution
OK, j'ai trouvé les deux solutions (x,y)=(4,9) et (9,4).
Même technique en procédant à deux changements de variables. La première, x-4=a, y-4=b, puis multiplication de tous les termes pas 3^a, mise en facteur pour obtenir une équation du second degré, et changement de variable à nouveau 3^a = A (A par exemple) et le tour est joué.
Sir factors are 243.81m and81m