SSC CHSL 5 Mar 2018 Shift 3 Paper

Expression : $9^{\sqrt{\mathit{a}}}+{40}^{\sqrt{\mathit{a}}}={41}^{\sqrt{\mathit{a}}}$
Since equation consists of natural numbers, then 'a' must be an integer. Thus, substituting 'a' by perfect square numbers :
Put a = 1
⇒ $9^{\sqrt{1}}+{40}^{\sqrt{1}}={41}^{\sqrt{1}}$
L.H.S. = $9+40=49\ne$ R.H.S.
Now,putting a = 4
⇒ $9^{\sqrt{4}}+{40}^{\sqrt{4}}={41}^{\sqrt{4}}$
L.H.S. = $9^2+{40}^2=81+1600=1681$
R.H.S. = ${41}^2=1681$
Thus, a = 4