Determine all possible non-negative integer solutions to the equation: 3^x + 5^y = 7^z + 1 [ Comment: * I have only been able to derive (x,y,z) = (0,0,0); (1,1,1) as two of the solutions to the given problem. * It can be shown that each of x, y and z must be odd ( whenever x,y and z are positive) and x=4p+1, y=4q+1 for non-negative integers p and q. * I have not been able to derive any other solution or able to conclusively prove the non-existance of any other solutions apart from the ones mentioned above. Proposed Methodology Considering x=0, we obtain; 5^y = 7^z; implying y=z=0. Similarly, considering y=0 and x=0 in turn, we respectively obtain, (x,z)=0 and (y,z) =(0,0) Hence, if at least one of x, y and z is 0; then, it follows that : (x,y,z) = (0,0,0) For z=1, we obtain; 3^x+ 5^y = 8; which is feasible only if : (x,y,z) = (1,1,1) Now, reducing both sides of the given equation to Mod 3; we obtain: 2 ^y = 2 ( Mod3 ).; so that y must be odd. so, 3^x + 5 = 7^z + 1 ( Mod 8). Since, 3^x = 1, 3(mod 8) and 7^z = 1,7 (mod 8); it follows that both x and z must be odd. Now, 3^x + 5^y = 7^z + 1 gives: 3^x + 5 = 7^y +1 (Mod 10). Since, both x and z are odd; this is possible if only if x = 1+4p and z = 1+4q for non-negative integers p and q. *** I am unable to proceed further. ** Consequently, I have not been able to derive solutions other than (0,0,0); (1,1,1) or to prove that no solutions other than these two satisfies the provisions of the problem ].