$(x+y)(x+z)(y+z) = 33...3$ (A 333 digit number that consists only of 3's)
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$\begingroup$ Is this a puzzle you made up yourself, or does it come from somewhere else? $\endgroup$– Gareth McCaughan ♦Jan 9, 2021 at 14:44
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$\begingroup$ Not that you need to fully factorize the RHS, but it's nothing more than 3 * (11..1) where the right term is $R_{333}$, the 333rd repunit. $\endgroup$– smciJan 10, 2021 at 8:24
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$\begingroup$ @xhienne 's variant "same question with (2x+y)(2y+z)(2z+x)" seems less non-trivial, because then we have to factorize $R_{333}$. Probabilistically we'd expect the answer to be "very unlikely". Even if we allowed negative integers. $\endgroup$– smciJan 10, 2021 at 8:28
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2 Answers
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The answer is
no
because
if the equation holds then the sum of the factors is 2(x+y+z), an even number. If three numbers have even sum then they cannot all be odd; but their product is an odd number, so they must all be odd. Contradiction.
answered Jan 9, 2021 at 12:51
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$\begingroup$ Brilliant! (for the story, I had started prime factorizing that huge 333-digit number) OK, now same question with (2x+y)(2y+z)(2z+x)... ;-) $\endgroup$– xhienneJan 9, 2021 at 13:20
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2$\begingroup$ @xhienne: you could ask that as a separate question, since it doesn't admit this solution. It will depend on the factorization of 33....3. Probabilistically we'd expect the answer to be "very unlikely". $\endgroup$– smciJan 10, 2021 at 8:19
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$\begingroup$ @smci That was intended as a joke, out of frustration of seeing such an elegant answer, because what you describe is exactly the way I started to solve this very question. $\endgroup$– xhienneJan 10, 2021 at 10:05
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$\begingroup$ Gareth, forgive me for asking, but that isn't the sum of all the factors, you've dropped quadratic crossterms? Vieta's equations are not applicable here. Unless I'm missing an extra line of reasoning, if so could you add it. $\endgroup$– smciJan 10, 2021 at 11:05
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$\begingroup$ No Vieta fanciness here. I just mean the sum of the three factors whose product is supposed to be $(10^333-1)/3$. $\endgroup$– Gareth McCaughan ♦Jan 10, 2021 at 12:03
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Answer:
No.
Explanation:
If there were such factors, then since $33\dots 3$ is odd, each of the factors $(x+y)$, $(x+z)$, and $(y+z)$ must also be odd. Since $x+y$ and $x+z$ have the same parity, by subtracting $x$ from both we see that $y$ and $z$ also have the same parity, hence $y+z$ is even, a contradiction.
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2$\begingroup$ At the time you posted this, there was already an answer (mine) saying the same thing with very similar reasoning, posted 14 hours earlier. Before posting an answer, please check whether someone else has already posted an equivalent one. Thanks! $\endgroup$– Gareth McCaughan ♦Jan 10, 2021 at 12:06