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Topic: Consecutive integer products (Read 356 times) |
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NickH
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Consecutive integer products
« on: Jun 7th, 2003, 3:16am » |
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Show that each of the following equations has no solution in integers x > 0, y > 0, n > 1.
1. x(x + 1) = yn
2. x(x + 1)(x + 2) = yn
3. x(x + 1)(x + 2)(x + 3) = yn
Does the pattern continue? (I don't know the answer to this one.)
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: Consecutive integer products
« Reply #1 on: Jun 7th, 2003, 6:38am » |
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The generalisation of this result is a very difficult problem. It took Erdos nine years to crack! You may like to check out this link: http://www.nrich.maths.org.uk/askedNRICH/edited/3945.html
I've made no significant advances with extending the particular cases to the general case, but I'll share it anyway...
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Obviously with the first case we know that HCF(x,x+1)=1, so clearly this cannot be a perfect power. That is, the highest factor of the LHS, x(x+1), will be 1, and the highest factor of the RHS, yn, will be y. Therefore no solutions exist for y>1 and clearly there are no solutions for y=1.
In a similar way, HCF(x,x+1,x+2)=2, so for y>2, x(x+1)(x+2)=yn will have no integral solutions. Checking the cases y=1,2, we confirm that there are no positive integer solutions at all. The same idea can be exteneded to x(x+1)(x+2)(x+3)=yn. The HCF of the LHS will be 3, so there will be no integral solutions for y>3. Checking, confirms that no solutions exist for y=1,2,3.
You can see (and which is why NickH probably stopped here), extending this to the general case by this method will fail. We can agree that for x(x+1)(x+2)...(x+k), the highest common factor will be k and so no solutions exist for y>k, however, it is not possible to confirm this with the cases y=1,2,3,...,k-1.
I feel that it'd be nice to use a different approach, so here are some of my (failed) ramblings with the first case.
As x and (x+1) are consecutive integers, their product, yn, must be even and as n>1, it must contain at least a factor of 4.
1+2+3+...+x=x(x+1)/2, so yn=2(1+2+...+x). As LHS has a factor of 4, the sum, 1+2+...+x, must be even. Therefore, if a solution exists, x must be odd.
We know that yn==0 mod 4. However, x(x+1)=x2+x and if x is odd, x2==1 mod 4, which means that x==3 mod 4; that is, x=4k–1.
That's it for the moment. I suspect in the end that it will be necessary to consider unique prime factorisation anyway.
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| « Last Edit: Jun 7th, 2003, 8:43am by Sir Col » |
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