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So, I was asked by my teacher in school to solve this problem it really had me stumped.The problem is as follows:Given that $x$ and $y$ are integers, which of the following cannot be expressed in the form $x^2+y^5$?

$1.)\ 59170$

$2.)\ 59012$

$3.)\ 59121$

$4.)\ 59149$

$5.)\ 59130$

Is it possible for an elegant solution and not tedious trial and error?

  • 0
    Was the teacher (by any chance) asking $x^2+y^2$ instead of $x^2+y^5$? I am not sure this is high-school level. – 2017-01-15
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    We may use modular arithmetic, but $mod 3, mod 4, mod 5$ doesn't work.. – 2017-01-15
  • 2
    If you have the right idea, you can see that $n \equiv 7 \pmod{11}$ cannot be written in the form $x^2 + y^5$ with integers $x,y$. – 2017-01-15
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    This would only exclude 3.) 59121. – 2017-01-15
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    If we see (3) and (4) are primes , can we proceed by applying theorem? – 2017-01-15

4 Answers 4

4

Since $9^5=59049$ the choice $y=9$ can obtain 1.), 4.), 5.) with $x=11,\,10,\,9$. Similarly, $x=162,\,y=8$ addresses 2).

Proving 3) is impossible requires only a modulo 11 analysis; note that $59121=7$. Squares are $0,1,4,9,5,3$. Since 11 is prime, nonzero fifth powers square to 1 by Fermat's Little Theorem, so fifth powers are $0,1,10$. Notice we can't get $7,18$ as sums.

  • 0
    Excellent job. I was about to write up similar based on @DanielFischer's comment. This is a much more elegant solution than brute force. – 2017-01-15
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    We hit the tape together. Nice work. – 2017-01-15
10

I think the answer is 59121. I just used the fact that the last digit of the fifth power of an integer from $0$ to $9$ is the the number itself. Also we have only $5$ digits to chose from the last digits of the square of a number ,viz, $0,1,4,5,6,9$. Now, for last digit to be 1 I will have to chose from square or fifth power of $10n$ ( for some natural number $n$) And $11$ or vice versa which is not possible for the given range of integers in the options.

Also,

a) the sum of fifth power of any of the positive integer with last digit as $6$ and square of any integer with last digit as $5$

And

b) sum of fifth power of any of the positive integer with last digit as $5$ and square of any integer with last digit as $4$ Is not equal to the given number with last digit as $1$ ,ie, $59121$

  • 1
    Why didn't you edit your old answer instead of positing another one? Also, $x$ or $y$ may be negative. – 2017-01-15
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    Sorry for the goof up, I am travelling by a train with a primitive phone and a bad internet connection. That's y this goof up. – 2017-01-15
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    +1 for the fact that last digit of the fifth power is number itself. – 2017-01-15
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    @Dietrich Burde : you are right my argument may hold true for positive integers only and I have not considered negative values for $x$ and $y$. – 2017-01-15
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    Won't this argument also work to show that $62661$ is not representable in the form $x^2+y^5$? – 2017-01-15
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    Zero is also a digit that can end a square number, which you do refer to obliquely later when selecting numbers to assess. Overall a nice filtering approach to cut down on a search – 2017-01-15
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    @naveendankal, but $62661=244^2+5^5$. – 2017-01-15
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    Yup I concurr with you and darn I just missed in those 4s and 5s as last digits – 2017-01-15
5

I have found $$59170=9^5+11^2$$ $$59012=8^5+162^2$$ $$59149=9^5+10^2$$ $$59130=9^5+9^2$$ and $59121$ can't be expressed in this form.

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    Which **is not** representable as $x^2+y^5$. I think you misread that part. – 2017-01-15
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    @SimpleArt But the answer also is useful, because we know that this number is not a candidate. – 2017-01-15
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    But we aren't supposed to use trial and error, which are pretty much the last 3 words. – 2017-01-15
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    @DietrichBurde Then it should be posted as a comment. This doesn't answer the question, even if we put aside the mis-reading. The question explicitly asks if there's a way to solve other than tedious trial and error. We could clearly find the answer by searching with a computer program in a couple of minutes; the question is asking for non-brute-force techniques. – 2017-01-15
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    @DietrichBurde No, I wouldn't change my mind. Not all questions have answers. (Especially since it's in the nature of mathematics that one can't very well post a rigorously justified answer of "There is no good method.") – 2017-01-15
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    @DavidRicherby In this case, if there is no good answer, one should at least not downvote the present answer. – 2017-01-15
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    @DietrichBurde No, it is a good question. Whether or not we are unable to answer a good question or not should not be reason to downvote the question. – 2017-01-15
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    @DietrichBurde I see no reason to downvote the question. The question is perfectly reasonable, even if it has no good answer. (And we don't know that it has no good answer; it's just that nobody's found one in the first half hour since it was posted.) – 2017-01-15
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    @DietrichBurde Actually, I see that a mod has posted what seems to be an elegant answer as a comment. (\*sigh\* Comments as answers, answers as comments, ...) – 2017-01-15
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    @DietrichBurde You are basically we should downvote the following 150 thousand questions should all be downvoted: http://math.stackexchange.com/questions?sort=unanswered – 2017-01-15
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    @DavidRicherby I do not want the question downvoted - I just found the critics to this answer too hard (and the $2$ downvotes it has). For those downvoters, please come up with a better solution. – 2017-01-15
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    you can try it for bigger numbers, or not? – 2017-01-15
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    @DietrichBurde No, I'm sorry. Me not having a better answer in no way precludes me from downvoting any other answer. By your argument, if I post "David is awesome" as an answer to any question that you can't solve, you're not allowed to downvote it. That's nonsense. – 2017-01-15
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    @DavidRicherby I fear this discussion will not enable me to make my point. I think the present answer is not so bad (of course perhaps not optimal), in any case better than "David is awesome". I cannot prevent you from downvoting, but I think it is not fair. – 2017-01-15
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    @DietrichBurde I agree that this post is much more constructive and useful than "David is awesome". It would make a helpful comment but it doesn't answer the question. – 2017-01-15
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    @DavidRicherby, you write "We could clearly find the answer by searching with a computer program in a couple of minutes." But the problem, as stated, is not restricted to *positive* integers $x$ and (especially) $y$. I don't see how simple brute force alone can rule out $59121$. – 2017-01-15
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    I would say this could at least qualify as a [partial answer](http://meta.math.stackexchange.com/questions/12718/what-to-do-with-partial-answer). As the OP did not mention knowing the correct solution, I would say that this at least answers *a part* of the question; and that the knowledge of the correct solution could set someone on the path to better, more complete answer. As this answer is neither [egregiously sloppy, a no-effort-expended post, or clearly and perhaps dangerously incorrect](http://stackoverflow.com/help/privileges/vote-down), I don't believe it deserves a downvote. – 2017-01-15
5

The use of $y^5$ makes me think that looking at the problem $\bmod 11$ - considering remainders from division by $11$ - would be useful.

Possible values of $x^2 \bmod 11 $ are $\{0,1,3,4,5,9\}$ and possible values of $y^5 \bmod 11 $ are $\{0,1,10\}\equiv \{-1,0,1\}$

The numbers given $(59170,59012,59121,59149,59130)$ have remainders of $(1,8,7,2,5)$, and the third one of these cannot be reached by the given expression $x^2+y^5$

So: if it is a single-answer question, the answer is certainly option (3).