What Are the Two Factors of 15120 Given Their Sum and Difference?

[RChristenk]

Homework Statement

Find two factors of $15120$ such that their difference is $6$ and their product is $15120$

Relevant Equations

Arithmetic

$15120=2^4 \cdot 3^3 \cdot 5 \cdot 7$. Now let the two factors be $a,b$. I am given $ab=15120$ and $a-b=6$.

$15120$ is an even number, hence one of the factors must be even. Since $a-b=6$, I can confirm both $a,b$ are even because only even subtracted by even results in even.

One of the factors must be divisible by $3$ because $15120$ itself is divisible by $3$. And then I'm stuck. How do I confirm the other factor is divisible or not divisible by $3$? I suspect it has something to do with $a-b=2\cdot3$, but I'm not sure how to prove it either way. Thanks.

 

[pasmith]

From $$(x + a)(x - b) = x^2 + (a - b)x - ab$$ we can see that $-a$ and $b$ are the roots of $$x^2 + 6x - 15120 = (x + 3)^2 - 15129 = (x + 3)^2 - 121^2.$$

 

[fresh_42]

Homework Statement: Find two factors of $15120$ such that their difference is $6$ and their product is $15120$
Relevant Equations: Arithmetic

$15120=2^4 \cdot 3^3 \cdot 5 \cdot 7$. Now let the two factors be $a,b$. I am given $ab=15120$ and $a-b=6$.

$15120$ is an even number, hence one of the factors must be even. Since $a-b=6$, I can confirm both $a,b$ are even because only even subtracted by even results in even.

One of the factors must be divisible by $3$ because $15120$ itself is divisible by $3$. And then I'm stuck. How do I confirm the other factor is divisible or not divisible by $3$? I suspect it has something to do with $a-b=2\cdot3$, but I'm not sure how to prove it either way. Thanks.

It is easier to solve the quadratic equation you get from Vieta's formulas $15120=a\cdot b = a\cdot (a-6)=a^2-6a,$ e.g. by completing the square, i.e. $0=a^2-6a-15120=(a-3)^2- 15129$ and compute the square root of $\sqrt{15129}=123,$ e.g. by using the Newton-Raphson method. This gives us $15120=120\cdot 126.$

Now, to your question: How to solve it by number theoretical methods?

The only instant tool I can think of is the definition of prime numbers: $p\,|\,(xy) \Longrightarrow p\,|\,x\text{ or } p\,|\,y.$ Since we know the prime factors of $15120$ we know that one of the factors $a,b$ must be divisible by $5$ and one by $7.$ You can therefore think of two urns $a,b$ filled with those primes $\{2,2,2,2,3,3,3,5,7\}.$ We know that each urn contains at least one $2$ and at least one $3:$
\begin{align*}
3\,|\,a\cdot b &\Longrightarrow \text{ (say) } 3\,|\,a\text{ (say) }a=3k\\
b=a-6=3(k-2) &\Longrightarrow 3\,|\,b
\end{align*}

That leaves you with possible urns $\{2,3,5,\ldots\}$ and $\{2,3,7,\ldots\}$ and since we know the solution, you can start seeking with the appropriate filling:
$$
\{a\}\cup \{b\}=\{2,2,2,3,5\}\cup\{2,3,3,7\}
$$
If you want to test all possibilities, then we have $\{2,2,2,3,3\}$ left to put in the two urns, i.e. $\binom 5 2 =10$ distributions to check.

 

[PeroK]

If $a - b = 6$, then $a$ and $b$ must be of a similar magnitude. A bit of trial and error should do it.

 

[PeroK]

Another idea is to take the square root of $15120$. The two factors must be approximately this $\pm 3$.

 

[SammyS]

From $$(x + a)(x - b) = x^2 + (a - b)x - ab$$ we can see that $-a$ and $b$ are the roots of $$x^2 + 6x - 15120 = (x + 3)^2 - 15129 = (x + 3)^2 - 121^2.$$

There is typo there (I hope). That should be:

$\displaystyle (x + 3)^2 - 15129 = (x + 3)^2 - 123^2$