Math Olympiad Problem | Solve a^x-b^x=369 | International Algebra Simplification | Find x, a & b ? 

Since you seem to assuming you're really working with integers you forgot a couple og solutions. We have 185^2 - 184^2 = 369 and 63^2 - 60^2 = 369. This covers the case x = 2.

See this answer 369=41×9=41×9×1=(5^2+4^2)(5+4)(5-4)=(5^2+4^2)(5^2-4^2) =(5^4-4^4) a=5,b=4,x=4 (a,b) can be (-5,-4) since exponent is even.

You do wrong,master

369= 123х3 Why is this option not considered? Why only 41 9?

One equation in three unknowns needs other side conditions to obtain only a finite number of solutions. Even assuming that a and b are integers is not enough. For example assume a and b are integers and x=1. Then a=b+369 has an infinite number of solutions for a and b.

625^1 - 256^1 =369, x=1........25^2 - 16^2 = 369, x=2

If you don’t indicate that a, b and x are integers, then for any real numbers b > 0 and x ≠ 0, you can solve for the value of a: a^x - b^x = 369 a^x = b^x + 369 a = (b^x + 369)^(1/x) and then the real numbers a = (b^x + 369)^(1/x) b>0 and x≠0 satisfy the equation. Assuming that a, b and x are integers, the procedure used to solve the equation obviously has errors because for example x = 2, a = ±25 and b = ±16 also turn out to be a solution to the problem. Also for x = 1 there are infinitely many trivial integer solutions. You really need to be more careful when posing the problem. When we leave solutions out, it means that the solution has procedural errors.

a,b,x are positive integers greater than 1.

The base is not equal so the power is not equal.

And other 2 solution: a=185;b=184;x=2 a=63;b=60;x=2.

369=369x1;369=123x3

Nice question!!!

a=+- 5i and b=+- 4i are also solutions