We can start by rewriting the equation as \(1 + 5^{t} = 2^{x} + 10*2^{y-1}*5^{z - 1}.\)

Now \(1 + 5^{t} \equiv 26 \pmod{100}.\) Also, \(2^{x} \equiv 6 \pmod{10}\) for \(x = 4m, m \ge 1.\) Further, \(16^{m} \pmod{100}\) cycles through \(16, 56, 96, 36, 76, 16, ....\), none of which is \(26.\) So for the RHS of the equation to have any chance of being equivalent to \(26 \pmod{100}\) we cannot have both \(y,z \ge 2,\) as this would make the second term of the RHS a multiple of \(100.\)

Now if \(y = z = 1\) then we have \(1 + 5^{t} = 16^{m} + 10.\) By observation \(t = 2, m = 1\) satisfies this equation, making \((x,y,z,t) = (4,1,1,2)\) our first solution. Noting the cycle of residues modulo \(100\) mentioned above we note that \(m = 1 + 5n\) for non-negative \(n\). So our equation in this case can be further refined to \(5^{t} = 16*16^{5n} + 9.\) It seems unlikely that there are any more solutions in this case, but I haven't come up with a proof yet.

If \(y = 1, z \gt 1\) then the equation becomes \(1 + 5^{t} = 16^{m} + 50*5^{z - 2}.\) For the RHS to be equivalent to \(26 \pmod{100}\) we will require that \(m = 5 + 5n\) for non-negative \(n.\) So our equation can be further refined to \(1 + 5^{t} = 16^5*16^{5n} + 50*5^{z-2}.\) Again, I need to work on a proof that there are no solutions.

If \(z = 1, y \gt 1\) then the equation becomes \(1 + 5^{t} = 16^{m} + 20*2^{y-2}.\) Since none of \(6, 86, 66, 46\) or \(26\) show up in the cycle of residues discussed above, there will be no solutions generated from this case.

So that narrows things down to two equations; others are welcome to continue along this train of thought and complete the proof. I need sleep. :)

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## Comments

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TopNewestJust a few thoughts to get the ball rolling ....

We can start by rewriting the equation as \(1 + 5^{t} = 2^{x} + 10*2^{y-1}*5^{z - 1}.\)

Now \(1 + 5^{t} \equiv 26 \pmod{100}.\) Also, \(2^{x} \equiv 6 \pmod{10}\) for \(x = 4m, m \ge 1.\) Further, \(16^{m} \pmod{100}\) cycles through \(16, 56, 96, 36, 76, 16, ....\), none of which is \(26.\) So for the RHS of the equation to have any chance of being equivalent to \(26 \pmod{100}\) we cannot have both \(y,z \ge 2,\) as this would make the second term of the RHS a multiple of \(100.\)

Now if \(y = z = 1\) then we have \(1 + 5^{t} = 16^{m} + 10.\) By observation \(t = 2, m = 1\) satisfies this equation, making \((x,y,z,t) = (4,1,1,2)\) our first solution. Noting the cycle of residues modulo \(100\) mentioned above we note that \(m = 1 + 5n\) for non-negative \(n\). So our equation in this case can be further refined to \(5^{t} = 16*16^{5n} + 9.\) It seems unlikely that there are any more solutions in this case, but I haven't come up with a proof yet.

If \(y = 1, z \gt 1\) then the equation becomes \(1 + 5^{t} = 16^{m} + 50*5^{z - 2}.\) For the RHS to be equivalent to \(26 \pmod{100}\) we will require that \(m = 5 + 5n\) for non-negative \(n.\) So our equation can be further refined to \(1 + 5^{t} = 16^5*16^{5n} + 50*5^{z-2}.\) Again, I need to work on a proof that there are no solutions.

If \(z = 1, y \gt 1\) then the equation becomes \(1 + 5^{t} = 16^{m} + 20*2^{y-2}.\) Since none of \(6, 86, 66, 46\) or \(26\) show up in the cycle of residues discussed above, there will be no solutions generated from this case.

So that narrows things down to two equations; others are welcome to continue along this train of thought and complete the proof. I need sleep. :)

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Adarsh Kumar Nihar Mahajan Lakshya Sinha @Dev Sharma

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