I tried to start this quest but won't end up with any answer. First I made a complete square on RHS which is like this $x^{2} - xy - xy + xy + y^{2}$ Then converted it into $xy+{\left( x - y\right) }^{2}$ And as LHS is a perfect square Which is $7^2$ So RHS must be a square but I don't know how to proceed further?
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If $x=y$, then $x=y=7$. So suppose $x>y$ (the other case is obviously symmetric). Notice that the equation can be rearranged as $x(x-y)=(7-y)(7+y)$, so $y<7$. Hence, you just have to check $6$ cases.
Leo163
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1Maybe add a sentence why the rearranged equation lets you conclude $y>7$? – quarague Jun 19 '19 at 13:59
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Oh, it's because the LHS is larger than $0$, so the factors in the RHS have to be either both positive or both negative: but they cannot be both negative, since $7+y>0$. So $7-y>0$, as we wanted. – Leo163 Jun 19 '19 at 14:02
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Put that comment into your answer and your solution will be even nicer :-) – quarague Jun 19 '19 at 14:27
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Thank you for your help – Ankit Kumar Jun 19 '19 at 15:28
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Do this. Multiply both sides by $4$: $$ x^2-xy+y^2=49\Rightarrow (2x-y)^2+3y^2 =196. $$ Now, this immediately yields, $y^2\leqslant \frac{196}{3}<66$, thus, $y\leqslant 8$. The rest is easy.
TBTD
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