We have LHS is multiple of 4 so RHS must be multiple of 4 or
b must be even and > 2^ a
For solution to exist we must have
4^a+4a^2+4 >= (2^a + 2)^2
Or 4^a+4a^2+4 >= 4^a + 2^(a+2) + 4
Or 4a^2 >= 4 * 2^a or a^2 >=2^ a
We see from it a <= 4. SO we need to try
for
A = 0 => b^2 = 5 no solution
A= 1 => b^2 = 2 no solution
A =2 => b = 6
A = 3=> b^2 = 64 + 36 +4 no solution
A= 4=> b^2 = 256+ 64 + 4 = 324 or b = 18
So solution set (2,6) and (4,18)
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