Question Number 191232 by BaliramKumar last updated on 21/Apr/23
$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2023}} \:\mathrm{then}\:\mathrm{how}\:\mathrm{many}\: \\ $$$$\mathrm{pair}\:\mathrm{of}\:{x},{y}\:{where}\:{x},\:{y}\:\in\:\mathrm{N} \\ $$
Answered by mr W last updated on 21/Apr/23
$$\mathrm{2023}=\mathrm{7}×\mathrm{17}^{\mathrm{2}} \\ $$$$\mathrm{2023}^{\mathrm{2023}} =\mathrm{7}^{\mathrm{2023}} ×\mathrm{17}^{\mathrm{4046}} \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)={a}×{b}=\mathrm{7}^{\mathrm{2023}} ×\mathrm{17}^{\mathrm{4046}} \\ $$$${x}−{y}={a} \\ $$$${x}+{y}={b}\:>\:{a} \\ $$$${x}=\frac{{a}+{b}}{\mathrm{2}}\:>{y} \\ $$$${y}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${a}\:{pair}\:{of}\:\left({a},\:{b}\right)\:\Rightarrow\:{a}\:{pair}\:{of}\:\left({x},\:{y}\right) \\ $$$${number}\:{of}\:{pairs}\:{of}\:\left({a},\:{b}\right)\:{is} \\ $$$$\frac{\left(\mathrm{2023}+\mathrm{1}\right)\left(\mathrm{4046}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{4}\:\mathrm{095}\:\mathrm{564} \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{4}\:\mathrm{095}\:\mathrm{564}\:{pairs} \\ $$$${of}\:{x},\:{y}. \\ $$
Commented by BaliramKumar last updated on 21/Apr/23
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$