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I have a question about number theory:

How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$

My attempt:

Move the $y^2$ from right to the left and 10! From left to the right such that:

$$x^2-y^2= 10!$$

$$(x-y)(x+y)=10!$$

$$(x-y)(x+y)= 10.9.8.7.6.5.4.3.2$$

Until this step, I think I will get many possibility to get the answer, such that:

$$(x-y)= 10.9$$ and $$(x+y)= 8!$$

Also

$$(x-y)= 10.9.8$$ and $$(x+y)= 7!$$

And another possibilities, like $$(x-y)=10.9.8.7$$ and $$(x+y)=6!$$

And so on until $$(x-y)=10.9.8.7.6.5.4.3.2$$ and $$(x+y)=1$$

And I think solving all the possibility is a bit tedious work. Can somebody help me to find a better solution to solve this kind of problem?

Thanks

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3 Answers 3

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$10!=2^83^45^27$
You want to divide this into two factors, one of which is a multiple of $7$.
The factors must both be even, otherwise $x$ and $y$ will not be integers.
So the factor that includes $7$ must have from one to seven 2's; from zero to four 3's and from zero to two 5's. How many options are there altogether?

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Write $10! = 10.9.8.7.6.5.4.3.2 =2^83^45^27 $.

You want $(x+y)(x-y)=10!=ab$ with $x+y=a$ and $x-y=b$. Then $x = (a+b)/2$ and $y = (a-b)/2$. So $a$ and $b$ have to have the same parity.

Then choose $2^a3^b5^c7^d$ with all possible choices satisfying the above. Since they have to have the same parity, $a$ must have $2^n$ where $1 \le n \le 6$.

This is a start.

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  • $\begingroup$ Oops - the discussion about same parity is wrong. $\endgroup$ Sep 22, 2015 at 14:44
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$105$ pairs.

As marty cohen said: "$(x+y)(x-y)=10!=ab$ with $x+y=a$ and $x-y=b$. Then $x = (a+b)/2$ and $y = (a-b)/2$. So $a$ and $b$ have to have the same parity."

$10!=2^{\color{red}{8 }}\cdot 3^{\color{red}{4}}\cdot 5^{\color{red}{2}} \cdot 7^{\color{red}{1}} \implies (\color{red}{8}+1)(\color{red}{4}+1)(\color{red}{2}+1)(\color{red}{1}+1)=270 \ $divisors in $10! \ $. Since $10!$ is an even number, each divisor must be even so that $x$ and $y$ will come out as integers when the divisors are added or subtracted and subsequently divided by $2$.

So we need

  1. all the ways $p=p_1\cdot p_2=3^4\cdot 5^2 \cdot 7^1$ can be split into two divisors, ... crossed with
  2. all the ways we can distribute the $2^8$ into two factors so that each factor has at least one $2$

For,

  1. $p=3^4\cdot 5^2 \cdot 7^1$ can be split into $\frac{270}{(\color{red}{8}+1)}=30$ different factors, that's $\frac{30}{2}=15$ unique choices of factor-pairs $p_1 \cdot p_2\ $ since the order of $p_1$ and $p_2$ doesn't change the outcome.

And for

  1. There's $(2^7p_1 \cdot 2^1p_2) \cdots (2^1p_1 \cdot 2^7p_2)$ for each choice of $p_1 \cdot p_2$, and I see $7$ of them.

So I see $7\cdot 15 =105$ different ways we can write $(x,y)$ down so that $x^2-y^2=10!$.

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