# Product of 25 consecutive integers can never be a perfect square

today exam is : Prove that the product of at most 25 consecutive integers is not a square.

this problem Also appears in this book： Problems from the Book proposed by Gabriel Dospinescu chapter 3.

I also know the general result which is 1939 Erdoshere 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.

Thanks.

• Have you thought about the prime factorisation? – jojobo Feb 23 at 7:56
• The product of zero items is conventionally $1$ and the product of one item which is a square is evidently a square, so these cases are presumably excluded. – Mark Bennet Feb 23 at 7:59

This rather elliptic argument (for twenty-five) is really mostly adapting Euler’s proof, which is extremely accessible (there’s no denying a high-schooler studying for the IMO could do that given enough time and perhaps a little encouragement). I admit not being too sure how to consider the case of less integers without redoing some 24 times the calculations.

Consider twenty-five consecutive integers the product of which is a square.

Define, for an integer $$n$$, the radical of $$n$$ to be the product of the prime divisors of $$n$$ who appear an odd number of times in the prime decomposition of $$n$$. We’ll consider the radicals of our $$25$$ consecutive integers (so they are square-free, and products of primrs less than $$25$$).

So, with elementary modular arithmetic, one finds that two appears in seven to ten radicals. Three appears in five to seven radicals. Five appears in four to five radicals. Seven appears in two to four radicals. Eleven appears in one to three radicals. Primes above eleven appear in at most two radicals.

Because the product of the consecutive integers is a square, two appears in eight or ten radicals, three in six, five in four, seven in two or four, eleven in two, primes above eleven in zero or two.

It follows that the product of the radicals divides $$N=2^{10}3^65^47^4(11\cdot 13 \cdot 17 \cdot 19 \cdot 23)^2$$.

But what are the first $$25$$ square-free numbers? They are $$1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23,26,29,30,31,33,34,35,37,38$$. Their product is $$M=2^93^65^57^411^313^217^219^2 \cdot 23 \cdot (29 \cdot 37)$$ so $$M > N$$.

It follows from this that two of the 25 numbers (let’s call them $$a < b$$) have the same radical $$r$$, and thus we can write $$a=rx^2,b=ry^2$$ and $$r(y^2-x^2) \leq 24$$. So $$r(2y+1) \leq 24$$, thus $$ry \leq 12$$, thus $$b \leq (ry)^2 \leq 144$$. We can check (counting powers of $$19$$ and $$23$$) that this is impossible.