In how many ways can 945 be written as difference of squares of two natural numbers ?

Question

anonymous · Answer

I got one way : $31^2 - 4^2$            :P

mathslover · Answer

5 = 6 - 1
5 = 7 - 2
5 = 8 - 3
5 = 9 - 4

So, two cases possible..
5 = 6 - 1 or 5 =  9- 4

$945 = (2k+1)^2 - (2k)^2 $
$945 = (2k+1-2k)(2k+1+2k)$
$945=(4k+1)$
$4k = 944$
$k = 236$ 

So, one case is this when k = 236 i.e the natural numbers are 472 and 473

ganeshie8 · Answer

i found 8 ways

mathslover · Answer

Now, let us consider that the two natural numbers are not consecutive. 

$945 = (2m + 1)^2 - (2n)^2 $ 
$945 = 4m^2 + 1 + 4m - 4n^2$
$945 = (4m^2 - 4n^2) + 4m + 1$
$944 = 4(m^2 - n^2 + m) \
236 = m^2 - n^2 + m $

mathmath333 · Answer

yes thats right , but how ?

mathmath333 · Answer

i m only concerned with the number of ways , the exact values are not relevant

mathslover · Answer

Oh, I see. Then there must be a specific algorithm for it. 

Would love to see how ganeshie approached the solution.

ganeshie8 · Answer

Look at prime factorization $$945 = 3^3\times 5\times 7 = x^2-y^2 = (x+y)(x-y)$$

mathslover · Answer

No, don't tell me that! Ganeshie... you just wasted a tree... here is how 

i) I saw the problem.
ii) I started working on it.
iii) There was no one else posting answer so I resumed my work.
iv) I finished a page of work. 
v) Then Ganeshie posted the solution.
vi) That one page got wasted.
vii) And in this way, one tree got consumed! :'(

Jokes apart, smartly done @ganeshie8

ganeshie8 · Answer

Its all good :) Own solution is always better than somebody else's solution :) it tells us alternative ways to look at the problem sometimes... but what i gave was just a hint...

ganeshie8 · Answer

$$\large 3^{\color{Red}{3}}\times 5^{\color{red}{1}}\times 7^{\color{red}{1}} =  (x+y)(x-y) $$
Next notice that $945$ has a total of $(\color{Red}{3}+1)(\color{Red}{1}+1)(\color{Red}{1}+1) = 16$ divosors. But we are only interested in positive solutions :  $x\gt 0 $ and $y \gt 0$

mathmath333 · Answer

perfect!

ganeshie8 · Answer

Suppose $x+y = d$ then we have $x-y = \dfrac{945}{d}$

solving gives 
$$x = \dfrac{d+\frac{945}{d}}{2}$$
$$y = \dfrac{d-\frac{945}{d}}{2}$$

ganeshie8 · Answer

Notice that $x$ is positive for all divosors $d$, but $y$ is positive only for half of the divisors. Consequently the number of solutions will be half of the total number of positive divisors of $945$

mathmath333 · Answer

thnks got it