This is a question from Bosnia and Hercegovina mathematical Olympiad $2002$.

There are already many similar questions, which I have done in the past.

For $5$ consecutive integers

A more general case for $3$ consecutive integers.

For $6$ consecutive integers

And similar proves are available for $3$ (Which was once asked in Korean Mathematical Olympiad) and $4$ (Indian national mathematical Olympiad) consecutive integers.

I also know the general result which is here but I am no way going to use that in a competition.

Please try to give a high school argument as supposed to be given in a national Olympiad.


  • $\begingroup$ I am still thinking about a basic proof. Maybe this will help: Multiplying 10 consecutive numbers will result in a number with last digit 0. Multiplying any number with itself ($x^2$) will have the last digit 0, only if x has the last digit 0. $\endgroup$ – P. Siehr Jul 17 '17 at 15:29
  • $\begingroup$ @P. Siehr Actually at least the last two digits will be $0$, but I am not sure it is a good approach to look only at the divisibility by $2$ and $5$ $\endgroup$ – charmd Jul 17 '17 at 15:31

Ah got it. Here is a Wordpress article which gives a proof which is not so simple but works fine at Olympiad level.

  • $\begingroup$ I was very close to solution $A9$ but before I could complete it you came up with the article. But the other solutions are also very beautiful. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:36
  • $\begingroup$ @астонвіллаолофмэллбэрг, did you use the same argument they are using? If not, then your answer will be gladly welcomed. $\endgroup$ – Vidyanshu Mishra Jul 17 '17 at 15:39
  • $\begingroup$ @VidyanshuMishra No, it's actually very similar, I don't want to post it. But I had something else in mind too, which I thought I could use differently. For example, consider the product of seven consecutive integers, which we can write as $(x-3)(x-2) (x-1)x(x+1)(x+2)(x+3)$. Now, expanding this out, and actually simplifying as a polynomial, we see this is equal to $x(x^2(x-7)^2 + 36)$. Now, if the product is a perfect square, then so is $x(x^2(x-7)^2 + 36)$. If $x$ is co - prime to $36$, then it's coprime to $x^2(x-7)^2 + 36$ as well. But then, each one will have to be a square. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:43
  • $\begingroup$ So either $x$ is not co prime to $36$, or there are two squares differing by $36$, wherein we know all the possibilities. But the question is, what if $x$ isn't co prime. There I was stuck (in the seven case) before I changed approach to counting primes in the interval $(x,x+10)$. Feel free to add thoughts, but I think this is a stuck point, although the polynomial observation is useful. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:44

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