This question arose when I was tackling another question, that there exists some function $f(x)=n^x$ such that if $a+b$ is a square, $f(a)+f(b)$ is also a square. I'm trying to disprove it by counter example, with the case $a=1, b=3$.
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So just to make things clear, your question is the title, correct? – Rick Oct 19 '15 at 00:43
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Yes, that is correct – Rob Oct 19 '15 at 00:49
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2If you are familiar with congruences, you can see that $9^x+1\equiv 2\pmod{4}$. But no square is congruent to $2$ modulo $4$. – André Nicolas Oct 19 '15 at 00:49
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@AndréNicolas Ooh, I like yours better. :) – Michael Biro Oct 19 '15 at 00:51
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To start, we can rewrite $9^x$ as $(3^2)^x$. Using the power rule, we can rewrite this as $3^{2x}$, and use the power rule again to get the expression $(3^x)^2$. This proves $9^x$ must be square.
Now that we know $9^x$ is square, we can use the obvious rule that a perfect square plus one is not a perfect square.
Meow Mix
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Suppose $9^x + 1 = y^2$ for $x > 0$, and split into two cases - if $y \leq 2$ or if $y > 2$.
If $y \leq 2$, we can quickly check that there are no solutions. If $y > 2$ then $9^x = y^2 - 1 = (y-1)(y+1)$. Since the only prime factor of $9^x$ is $3$, $3$ must divide $y - 1$. Then $3$ does not divide $y + 1$, a contradiction.
Michael Biro
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