Question Number 161843 by Rasheed.Sindhi last updated on 23/Dec/21
$${Q}#\mathrm{161744}\:{reposted}\:{with}\:{some}\:{change}. \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\boldsymbol{\mathrm{integer}}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\:+\:\frac{\mathrm{y}\:-\:\mathrm{5}}{\mathrm{5}}\:=\:\frac{\mathrm{y}\:+\:\mathrm{x}}{\mathrm{y}\:+\:\mathrm{5}}\:+\:\frac{\mathrm{5}\:+\:\mathrm{y}}{\mathrm{5}\:+\:\mathrm{x}} \\ $$
Commented by malwan last updated on 23/Dec/21
$${x}=\mathrm{0}\:{or}\:{x}=\mathrm{5} \\ $$$${then}\:{we}\:{have}\:{one}\:{equation} \\ $$$${and}\:{one}\:{variable}\:{y} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Dec/21
$${x}\neq\mathrm{0}\:{because}\:{it}'{s}\:{in}\:{denominator}. \\ $$$${See}\:{Q}#\mathrm{161744} \\ $$
Commented by mr W last updated on 23/Dec/21
$${x}\neq\mathrm{0},\:{x}\neq−\mathrm{5},\:{y}\neq\mathrm{0},\:{y}\neq−\mathrm{5} \\ $$$${we}\:{can}\:{see}\:{x},\:{y}\:{could}\:\left({but}\:{not}\:{must}\right) \\ $$$${be}\:{multiples}\:{of}\:\mathrm{5},\:{then}\:{we}\:{can}\:{get} \\ $$$${follwing}\:{solutions}: \\ $$$$\left(\mathrm{5},\mathrm{5}\right) \\ $$$$\left(\mathrm{5},\mathrm{10}\right) \\ $$$$\left(−\mathrm{10},\mathrm{5}\right) \\ $$$${but}\:{it}'{s}\:{not}\:{easy}\:{to}\:{prove}\:{that}\:{other} \\ $$$$\left({not}\:{multiples}\:{of}\:\mathrm{5}\right)\:{solutions}\:{don}'{t} \\ $$$${exist}. \\ $$