A problematic problem

Number Theory Level 4

$\large xy=2^2\cdot3^4\cdot5^7(x+y)$

If $x$ and $y$ satisfy the above condition, then find the number of positive integral solution(s).

200 675 0 Infinite

2 solutions

Kushal Bose
Jul 3, 2016

$xy=2^{2}3^{4}5^{7}(x+y)$

Let $2^{2}3^{4}5^{7}=p$ So, $xy=px + py$

$xy-px-py + p^{2}= p^{2}$

$(x-p)(y-p)=p^{2}$

So, $(x-p)(y-p)=2^{4}3^{8}5^{14}$

Here the number factors of the right side is number of positive solutions in x and y

So number of solutions are $(4+1)(8+1)(14+1)=675$

Kalpok Guha
Jul 2, 2016

We have $xy=2^2\cdot3^4\cdot5^7(x+y)$

or $\frac{x+y}{xy}=\frac{1}{2^2\cdot3^4\cdot5^7}$

or. $\frac{1}{x}+\frac{1}{y}=\frac{1}{2^2\cdot3^4\cdot5^7}$

By S.F.F.T ( Simon's Favorite Factoring Trick )

$(2^2\cdot3^4\cdot5^7-x)(2^2\cdot3^4\cdot5^7-y)=(2^2\cdot3^4\cdot5^7)^2=2^4\cdot3^8\cdot5^{14}$

Number of integral solutions of this equation is $(4+1)(8+1)(14+1)=5.9.15=\boxed{675}$

@Chirayu Bhardwaj : Since $x$ and $y$ are integers, the number of solutions is actually $2 \cdot 5 \cdot 9 \cdot 15 = 1350$ . (because the factors $2^2 \cdot 3^4 \cdot 5^7 - x$ and $2^2 \cdot 3^4 \cdot 5^7 - y$ can be positive or negative).

Jon Haussmann - 4 years, 11 months ago

Sir i think it is only valid in some cases where one of the integers is negative and it is greater than the positive one . Not valid for both the integers negative . Well i have made changes in the question :) . I hope the query is solved.

Chirayu Bhardwaj - 4 years, 11 months ago

A problematic problem

2 solutions

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