perplexus.info :: Just Math : So many pairs...

So many pairs... (Posted on 2023-08-24)

Please find all pairs (x, y) satisfying the equation:
16^(x^2+y)+16^(y^2+x)=1

No Solution Yet Submitted by Ady TZIDON

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I found one pair | Comment 1 of 3

Each parenthetical expression could be equal to -1/4;
Because 16^(-1/4) = 1/2

x^2+y = y^2+x = -1/4
x^2+y + y^2+x = -1/2
x^2+x+1/4 + y^2+y+1/4 = 0
(x + 1/2)^2 + (y + 1/2)^2 = 0
x = y = -1/2

(x,y) = (-1/2, -1/2)

Posted by Larry on 2023-08-24 17:14:27

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