2^x +9^y =x^2 +9xy+y^2 Find x,y∈N


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Question Number 143706 by Huy last updated on 17/Jun/21

$2^{x} + 9^{y} = x^{2} + 9 xy + y^{2}$ $Find x, y \in N$
	Answered by TheHoneyCat last updated on 17/Jun/21
	$for x=10 we get: 1024+9^y =^? 100+90y+y^2 checking in y=10 and studing the derivative one gets: ∀y≥10, 1024+9^y >100+90y+y^2 looking at how the values evolve for x (the derivative for x) we get also that ∀x,y≥10 2^x +9^y >x^2 +9xy+y^2 Checking, with a computer, for all the values (x,y)∈[∣0,10∣]^2 I found that there are only two possibilites: (x,y)∈{(1,1);(3,0)} by the way, it is surely possible to get a better maximum value than ′′10′′ so that the verification can be done by hand I was just too lazy to do it efficiently$
	$for x = 10 we get :$ $1024 + 9^{y} \overset{?}{=} 100 + 90 y + y^{2}$ $checking in y = 10 and studing the derivative$ $one gets :$ $\forall y ⩾ 10, 1024 + 9^{y} > 100 + 90 y + y^{2}$ $looking at how the values evolve for x (t h e d e r i v a t i v e f o r x)$ $we get also that \forall x, y ⩾ 10 2^{x} + 9^{y} > x^{2} + 9 x y + y^{2}$ $Checking, with a computer, for all the values (x, y) \in {[∣ 0, 10 ∣]}^{2}$ $I found that there are only two possibilites :$ $(x, y) \in {(1, 1); (3, 0)}$ $b y t h e w a y, i t i s s u r e l y p o s s i b l e t o g e t a b e t t e r m a x i m u m v a l u e t h a n^{″} 10^{″}$ $s o t h a t t h e v e r i f i c a t i o n c a n b e d o n e b y h a n d$ $I w a s j u s t t o o l a z y t o d o i t e f f i c i e n t l y$
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