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Question Number 31763 by RAMANUJAN last updated on 14/Mar/18

please find the integral solutions (x and y) (xy−7)^2 =x^2 +y^2

$${please}\:{find}\:{the}\:{integral}\:{solutions}\:\left({x}\:{and}\:{y}\right)\: \\ $$$$\left({xy}−\mathrm{7}\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \\ $$

Answered by MJS last updated on 14/Mar/18

(−7,0) (−4,−3) (−3,−4) (0,−7) (0,7) (3,4) (4,3) (7,0)

$$\left(−\mathrm{7},\mathrm{0}\right) \\ $$$$\left(−\mathrm{4},−\mathrm{3}\right) \\ $$$$\left(−\mathrm{3},−\mathrm{4}\right) \\ $$$$\left(\mathrm{0},−\mathrm{7}\right) \\ $$$$\left(\mathrm{0},\mathrm{7}\right) \\ $$$$\left(\mathrm{3},\mathrm{4}\right) \\ $$$$\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\left(\mathrm{7},\mathrm{0}\right) \\ $$

Commented by MJS last updated on 14/Mar/18

...now I see that 4×3−7=5 4^2 −3^2 =7 3^2 +4^2 =5^2 these might have helped with the construction of the given equation

$$...\mathrm{now}\:\mathrm{I}\:\mathrm{see}\:\mathrm{that} \\ $$$$\mathrm{4}×\mathrm{3}−\mathrm{7}=\mathrm{5} \\ $$$$\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} =\mathrm{7} \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{these}\:\mathrm{might}\:\mathrm{have}\:\mathrm{helped}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{construction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation} \\ $$

Commented by Joel578 last updated on 14/Mar/18

Sir pls explain the way. I′m just guessing the answer

$$\mathrm{Sir}\:\mathrm{pls}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{way}.\:\mathrm{I}'\mathrm{m}\:\mathrm{just}\:\mathrm{guessing}\:\mathrm{the} \\ $$$$\mathrm{answer} \\ $$

Commented by MJS last updated on 14/Mar/18

I solved the equation y=((7x±(√(x^4 −x^2 +49)))/(x^2 −1)) ⇒ x^4 −x^2 +49 must be a square number then I tried...

$$\mathrm{I}\:\mathrm{solved}\:\mathrm{the}\:\mathrm{equation} \\ $$$${y}=\frac{\mathrm{7}{x}\pm\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{49}}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\:{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{49}\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square}\:\mathrm{number} \\ $$$$\mathrm{then}\:\mathrm{I}\:\mathrm{tried}... \\ $$

Answered by Joel578 last updated on 14/Mar/18

(0, ±7) and (±7, 0)

$$\left(\mathrm{0},\:\pm\mathrm{7}\right)\:\mathrm{and}\:\left(\pm\mathrm{7},\:\mathrm{0}\right) \\ $$

Answered by ajfour last updated on 14/Mar/18

x^2 y^2 −14xy+49=x^2 +y^2 x^2 y^2 −16xy+64=(x−y)^2 +15 (xy−8)^2 =(x−y)^2 +15 (xy−8−x+y)(xy−8+x−y)=15 ⇒ xy−8−x+y=3 and xy−8+x−y=5 ⇒ xy=12 and x−y=1 ⇒ (4,3) or (−3,−4) alternatively xy−8−x+y=5 and xy−8+x−y=3 ⇒ xy=12 and y−x=1 so (3,4) or (−4,−3) alternatively xy−8−x+y=−5 and xy−8+x−y=−3 ⇒ xy=8 and x−y=1 ⇒ no integral solution or xy−8−x+y=−3 and xy−8+x−y=−5 ⇒ xy=8 and y−x=1 ⇒ no integral solution , again. otherwise xy−8−x+y =15 and xy−8+x−y=1 ⇒ xy=16 and y−x=7 ⇒ no integral solution alternatively xy−8−x+y=−15 and xy−8+x−y=−1 ⇒ xy=0 and x−y=7 ⇒ (7,0) and (0,−7) otherwise xy−8−x+y=−1 and xy−8+x−y=−15 ⇒ xy=0 and y−x=7 ⇒ (0,7) and (−7,0) hence the integral solutions for (x,y) are (4,3), (−3,−4), (3,4), (−4,−3), and (7,0), (0,−7), (0,7), (−7,0) .

$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{14}{xy}+\mathrm{49}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{16}{xy}+\mathrm{64}=\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{15} \\ $$$$\left({xy}−\mathrm{8}\right)^{\mathrm{2}} =\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{15} \\ $$$$\left({xy}−\mathrm{8}−{x}+{y}\right)\left({xy}−\mathrm{8}+{x}−{y}\right)=\mathrm{15} \\ $$$$\Rightarrow\:{xy}−\mathrm{8}−{x}+{y}=\mathrm{3}\:\:\:{and} \\ $$$$\:\:\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=\mathrm{5} \\ $$$$\Rightarrow\:\:{xy}=\mathrm{12}\:\:{and}\:\:{x}−{y}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{4},\mathrm{3}\right)\:{or}\:\left(−\mathrm{3},−\mathrm{4}\right) \\ $$$${alternatively} \\ $$$$\:\:\:\:\:\:\:{xy}−\mathrm{8}−{x}+{y}=\mathrm{5}\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=\mathrm{3} \\ $$$$\Rightarrow\:\:\:{xy}=\mathrm{12}\:\:{and}\:\:{y}−{x}=\mathrm{1} \\ $$$${so}\:\:\left(\mathrm{3},\mathrm{4}\right)\:{or}\:\left(−\mathrm{4},−\mathrm{3}\right) \\ $$$${alternatively} \\ $$$$\:\:\:\:\:\:\:\:\:{xy}−\mathrm{8}−{x}+{y}=−\mathrm{5}\:\:{and} \\ $$$$\:\:\:\:\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=−\mathrm{3} \\ $$$$\Rightarrow\:\:\:\:\:{xy}=\mathrm{8}\:\:\:{and}\:\:\:{x}−{y}=\mathrm{1} \\ $$$$\Rightarrow\:\:{no}\:{integral}\:{solution} \\ $$$${or}\:\:\:\:\:{xy}−\mathrm{8}−{x}+{y}=−\mathrm{3}\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=−\mathrm{5} \\ $$$$\Rightarrow\:\:\:\:\:{xy}=\mathrm{8}\:\:{and}\:\:{y}−{x}=\mathrm{1} \\ $$$$\Rightarrow\:{no}\:{integral}\:{solution}\:,\:{again}. \\ $$$${otherwise} \\ $$$$\:\:\:\:{xy}−\mathrm{8}−{x}+{y}\:=\mathrm{15}\:\:{and} \\ $$$$\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=\mathrm{1} \\ $$$$\Rightarrow\:{xy}=\mathrm{16}\:\:{and}\:{y}−{x}=\mathrm{7} \\ $$$$\:\:\:\:\Rightarrow\:{no}\:{integral}\:{solution} \\ $$$${alternatively} \\ $$$$\:\:\:\:{xy}−\mathrm{8}−{x}+{y}=−\mathrm{15}\:\:{and} \\ $$$$\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{xy}=\mathrm{0}\:\:\:{and}\:\:{x}−{y}=\mathrm{7} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{7},\mathrm{0}\right)\:\:{and}\:\left(\mathrm{0},−\mathrm{7}\right) \\ $$$$\:{otherwise} \\ $$$$\:\:\:\:\:\:{xy}−\mathrm{8}−{x}+{y}=−\mathrm{1}\:\:{and} \\ $$$$\:\:\:\:\:\:{xy}−\mathrm{8}+{x}−{y}=−\mathrm{15} \\ $$$$\Rightarrow\:\:\:\:{xy}=\mathrm{0}\:\:{and}\:\:{y}−{x}=\mathrm{7} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{0},\mathrm{7}\right)\:\:{and}\:\:\left(−\mathrm{7},\mathrm{0}\right) \\ $$$${hence}\:{the}\:{integral}\:{solutions} \\ $$$${for}\:\left({x},{y}\right)\:{are}\:\: \\ $$$$\:\left(\mathrm{4},\mathrm{3}\right),\:\left(−\mathrm{3},−\mathrm{4}\right),\:\left(\mathrm{3},\mathrm{4}\right),\:\left(−\mathrm{4},−\mathrm{3}\right), \\ $$$${and}\:\left(\mathrm{7},\mathrm{0}\right),\:\left(\mathrm{0},−\mathrm{7}\right),\:\left(\mathrm{0},\mathrm{7}\right),\:\left(−\mathrm{7},\mathrm{0}\right)\:. \\ $$

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