Happy new year.
Can we find $x,y \in \mathbb{N} $ such that : $$x^y -y^x=2020$$
One solution is $(x,y) = (2021,1)$. It is clear that both of $x,y$ are odd or even .
Note that $2020=2^2 \times 5 \times 101 $
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The Demonix _ Hermit
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amir bahadory
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9$2021^1 - 1^{2021} = 2020$ – iamwhoiam Jan 02 '20 at 10:19
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@iamwhoiam. Yes . Is answer unique? – amir bahadory Jan 02 '20 at 10:21
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1I was merely suggesting that the question as stated is asking one to prove a false statement. – iamwhoiam Jan 02 '20 at 10:23
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1Compare with $x^y-y^x=0$. – Dietrich Burde Jan 02 '20 at 10:29
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2What have you attempted ? The case when $x,y$ are both even is fairly simple to handle. – The Demonix _ Hermit Jan 02 '20 at 10:35
1 Answers
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If $x$ or $y$ is $1$, then $y=1$ and $x=2021$.
If $x$ or $y$ is even then both are even. Since $2020$ is not divisible by $8$, either $x$ or $y$ is $2$. For $ t\in \mathbb{N}$,$2^t-t^2\ge 0 $ if $t\ne 3$ and is an increasing function of $t$ for $t\ge 4$. Therefore $x=2$ and $y\approx 11$ but $2^{11}-11^2=1927\ne 2020. $
We now have $x,y\ge 3$ and therefore $x^y>y^x$ only for $y > x$.Then $$2020=x^x\left(x^{y-x}-\left(1+\frac{y-x}{x}\right)^x\right)\ge x^x\left(x^{y-x}-e^{y-x}\right) $$ and so $x<5$.
Therefore $x=3$ and $$2020\ge 27\left(3^{y-3}-e^{y-3}\right) $$ then $y\le 7$ and the values $y=5$ and $7$ are easily checked.