Kwesi's integers

Rewrite the expression as $y^2-x^2=10!$ and factorize into $(x-y)(x+y)=2^83^45^27$ For $x$ and $y$ to be integers we must have the parity of $(x+y)$ and $(x-y)$ be the same. Since at least one has a factor of $2$ , they both have a factor of $2$ . We must then distribute the remaining factors of $2, 3, 5,$ and $7$ among $(x+y)$ and $(x-y)$ .

We have $2^63^45^27$ left after giving each term a factor of $2$ . Each term can have between $0$ and $6$ factors of $2$ , $0$ and $4$ factors of $3$ , $0$ and $2$ factors of $5$ , and $0$ and $1$ factors of $7$ , so there are in total $7\times5\times3\times2=210$ factors to distribute.

We must have $x$ and $y$ positive, so we must have $(x+y)\geq(x-y)$ . Since $2^63^45^27$ is not a perfect square, for every factor $a$ there is an unequal factor $b$ such that $ab=2^63^45^27$ , there are $\frac{210}{2}=105$ such pairings $(a,b)$ and for each pairing we assign the greater factor to $(x+y)$ Each assignment is valid and gives unique values $x$ and $y$ , so the number of ordered pairs $(x,y)=\boxed{105}$

Shoutouts to Kwesi for the problem

$10!=(y-x)(y+x)$ So we have two factors, both even because we cannot have odd x even or odd x odd. $10!=2^{8}\cdot{3^4}\cdot{5^2}\cdot{7}$ $y-x=2^a\cdot{d_1}$ and $y+x=2^b\cdot{d_2}$ where $a,b>0$ and $d_1,d_2$ are divisors of $\frac{10!}{2^8}$

How many divisors?: $5\cdot{3}\cdot{2}$ (the exponents are $4,2,1$ )

In how many ways can we choose $a$ ?: $7$ because $0<a<8$

The solution is $\frac{7\cdot{5}\cdot{3}\cdot{2}}{2}=\boxed{105}$ because $y>x$ .

Rearrange the equation we will get:

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

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

Obviously, $10!$ is an even number, then $(y+x)$ and $(y-x)$ both will also be even numbers

$10!=2^8\times3^4\times5^2\times7$

The number of solutions to express $10!$ as a product of two EVEN numbers (including symmetries, for example: $2\times3$ and $3\times2$ are counted as two solutions) are

$S_{even}(10!)=7\times5\times3\times2=210$

7 is used instead of 9 because:

1) The number must be even, so $2^0$ is removed.

2) The other number must be even too! Remember to share, so another $2$ is removed

Next, we handle about the symmetry problem. $(x+y)$ and $(x-y)$ will not be symmetries as both $x$ and $y$ are positive integers in this case, so the number of solutions (that included symmetry) should be divide by 2, which left $\frac{210}{2}=105$ solutions.

Now, we have 105 ways to express $10!$ as a product of two even numbers (excluded symmetries). For each ways, there exist only 1 ordered pair of $(x,y)$ .

Thus, we get the answer $\boxed{105}$ .

The sketch of my solution

• make the expression easy to handle with
• reduce the problem to a fairly easier one

The $x^{2}$ and $10!$ doesn't look good together so first I shift it to the R.H.S. which I can then factorize.

Now we have, $10! = (y+x)(y-x)$

Now I'd write 10! as $2^{8} \times 3^4 \times 5^2 \times 7$ so that I can clearly see what is really going on. (It's is fairly easy to prime factorize 10!)

Thus we have, $2^{8} \times 3^4 \times 5^2 \times 7 = (y+x)(y-x)$

We are now done with simplifying the expression.

Now $y+x$ and $y-x$ are both integers since y,x are both integers, Assume that $y+x=a$ and $y-x=b$ ,[notice that a>b]

Solving for y we get, $y= \frac{a+b}{2}$

Since y is integer then a,b must be of same parity. Both a,b can't be simultaneously odd since $a \times b = 10!$

Thus a,b both are even.

(I could choose to solve for x but that'd have given us the same information,ie. a and b are even)

Now our problem is reduced to determine the number of ways to distribute the prime factors $2^{8} \times 3^4 \times 5^2 \times 7$ to two even parts.

Since both a,b are even we first give them what they crave for, ie. the factor 2 .

Now we are left with $2^{8-2} \times 3^4 \times 5^2 \times 7$ = $2^{6} \times 3^4 \times 5^2 \times 7$

Whenever I break this number into two factors,we'd get two different numbers since there is only one 7 in it.

Thus whenever I break the number I'd assign the larger number to $a$ and the smaller one to $b$ since a>b.

The problem is simply reduced to this--

In how many ways can $2^{6} \times 3^4 \times 5^2 \times 7$ be written as product of two numbers where $a \times b$ is the same as $b \times a$ . ?

This is really easy to solve,we count the number of factors of $2^{8} \times 3^4 \times 5^2 \times 7$ and divide by 2 as we counted each pair twice.

Thus we get the answer $\boxed {105}$

First of all:

$10! = y^2 - x^2 = (y + x)(y - x)$ .

We can carefully factorize $10!$ to $2^8 * 3^4 * 5^2 * 7$ . If we want to find the quantity of divisors of this, we know that for number $2$ we have $8$ diferent values (as we have $2^8$ ), $4$ different ones for $3$ , $2$ for $5$ and $1$ for $7$ . But we need all the divisors, so every quantity of prime factors counts. We make the sum $8 + 4 + 2 + 1 + 8*4 + 8*2 + 8*1 + 4*2 + 4*1 + 2*1 + 8*4*2 + 8*4*1 + 8*2*1 + 4*2*1 + 8*4*2*1$ (we can do it manually and carefully), and this gives us a total of $269$ . We sum $1$ more, because we need to include the $1$ as a factor, and $10!$ has $270$ divisors.

Now, note that $10!$ is even, and it is expressed as a square difference. We know that $(y +x) - (y - x)$ is of the form $2k$ (the difference is even). So, we need to exclude the odd factors, because they are forced to be accompanied with an even factor (because odd * odd = odd, and $10!$ is even). The odd factors, clearly, are among $3^4 * 5^2 * 7$ . With the same idea we found the total of divisors of $10!$ , we find that the odd divisors (including $1$ ) are a total of $30$ . As all of them will be accompanied with an specific factor, we need to exclude $30*2 = 60$ factors from the $270$ we've calculated. So, we get $210$ factors that can be expressed as a difference of squares with a specific partner. As we are looking for the pairs, the final answer is $210/2$ = $\boxed{105}$ .