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# RMO Practice!

Find the number of positive integers $$x,y,z,t$$ such that $1 + 5^{t} = 2^{x} + 2^{y} 5^{z}$.

Note by Harsh Shrivastava
2 years, 4 months ago

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Just 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. :)

- 2 years, 4 months ago

Adarsh Kumar Nihar Mahajan Lakshya Sinha @Dev Sharma

- 2 years, 4 months ago