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PostPosted: Sun, 31 May 2009 00:39:38 UTC 
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Prove that the product of five consecutive integers cannot be a perfect square.


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PostPosted: Sun, 31 May 2009 04:06:25 UTC 
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mathemagics wrote:
Prove that the product of five consecutive integers cannot be a perfect square.


Easy case of Erd"{o}s-Selfridge.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 2 Jun 2009 07:18:22 UTC 
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If you allow zero, this is not true.

The product of 5 consecutive POSITIVE integers can have only one 5 in its prime factorization...


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PostPosted: Tue, 2 Jun 2009 08:48:41 UTC 
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Quote:
The product of 5 consecutive POSITIVE integers can have only one 5 in its prime factorization...


Not true: 25 \cdot 26\cdot 27\cdot 28\cdot 29 has two 5's...

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PostPosted: Sat, 13 Jun 2009 20:20:02 UTC 
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Quite correct.

In fact, if we assume the product to be a perfect square, any prime > 3 that is a factor of any of the five integers must be a factor of only one of the integers and must have an even power, as your example demonstrates.

So each of the integers, in its prime factorization, can have only 2 or 3 with an odd power, and so each is one of the following forms:

m^2, 2m^2, 3m^2, 6m^2

Since there are five integers, two of the integers have one of these forms.
Let's go case by case with these forms.

The only squares close enough together to fit within five consecutive integers are 1 and 4, but 1*2*3*4*5 = 120 is not a perfect square. This eliminates the first form as a possibilty.

If two of the integers have the form 2m^2, then their difference being 4 or less impies that there are two squares with a difference of 2 or less. But no two squares exist with a difference of 1 or 2. This eliminates the second form as a possibilty.

Similarly, if two of the integers have the form 3m^2, then their difference of less than 5 implies there are two squares with a difference of 1, which again is impossible. This eliminates the third form as a possibilty.

Of course, there cannot be two multiples of 6 within any five consecutive integers, so the fourth form is also impossible.

Since all the possible forms are eliminated, the product of five consecutive integers cannot be a perfect square.


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