RMO Practice!

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

#Algebra #NumberTheory #ModularArithmetic #OrderedPairs

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

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. :)

Brian Charlesworth - 5 years, 6 months ago

Adarsh Kumar Nihar Mahajan Lakshya Sinha @Dev Sharma

Harsh Shrivastava - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$