VicKmath7 wrote: Find all positive integers $ a, b, c, d $ so that $ a^2+b^2+c^2+d^2=13.4^n $ For $n\ge 2$ note that $\text{RHS}\equiv 0\mod 8$ hence all $a,b,c,d$ are even.Let $(a,b,c,d)=(2\cdot a,'2\cdot b,'2\cdot c',2\cdot ,d')$.Then we get $(a')^2+(b')^2+(c')^2+(d')^2=13.4^{n-1}$.Hence $n$ is continuously decreasing till $n=1$.Hence we get $(a*)^2+(b*)^2+(c*)^2+(d*)^2=13.4=52$.Its easy to check that the solution for this are $(7,1,1,1),(4,4,4,2),(5,5,1,1)$.SO the solutions are $(2^{n-1}\cdot 7,2^{n-1}\cdot 1,2^{n-1}\cdot 1,2^{n-1}\cdot 1),(2^{n-1}\cdot 4,2^{n-1}\cdot 4,2^{n-1}\cdot 4,2^{n-1}\cdot 2),(2^{n-1}\cdot 5,2^{n-1}\cdot 5,2^{n-1}\cdot 1,2^{n-1}\cdot 1)$