# Exponential Diophantine perfect square problem

I need to find all positive integer solutions $(a,b,c)$ to $1+4^a+4^b=c^2$. I am certain that the only solutions are $(t,2t-1, 2^{2t-1}+1)$ and $(2t-1, t, 2^{2t-1}+1)$ , $t\in \mathbb{N}$ but I am having some trouble confirming this. Could someone help me please? Thank you.

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Sorry, I should have been more clear. I need help in providing a mathematical proof that those are indeed the only solutions. –  Bernhard Freeman Aug 12 '11 at 2:08

By symmetry it is enough to deal with the case $1 \le a \le b$. Rewrite the equation as $$4^a(1+4^{b-a})=c^2-1=(c-1)(c+1).$$

Suppose first that $b=a$. Then $4^a(1+4^{a-b})=2^{2a+1}$. But $(c-1)(c+1)$ is a power of $2$ only if $c=3$. Thus $a=b=1$. From now on we can assume that $b>a$.

Since $a \ge 1$, $c$ must be odd, say $c=2d+1$. Substitute and divide by $4$. We get $$4^{a-1}(1+4^{b-a})=d(d+1).$$ We cannot have $a=1$, since if $a=1$ the left side is odd and the right side is even. Thus $a>1$.

Since $b>a$, $1+4^{b-a}$ is odd. But one of $d$ and $d+1$ is even. Whichever one this is, it must be divisible by $4^{a-1}$. We first examine the possibility that $d$ or $d+1$ is exactly $4^{a-1}$.

Then either (i) $4^{a-1}=d$ and $4^{b-a}+1=d+1$, or (ii) $4^{a-1}=d+1$ and $4^{b-a}+1=d$.

Possibility (ii) is easy to dismiss: two non-trivial powers of $4$ cannot differ by $2$.

If possibility (i) holds, we have $b-a=a-1$, so $b=2a-1$. It is easy to verify that this always does give a solution of the original system, namely your solutions with $t>1$.

Finishing the argument: We found that one of $d$ or $d+1$ is divisible by $4^{a-1}$, and dealt fully with the case of equality. So now we ask: could we have $d$ or $d+1$ equal to $4^{a-1}k$, where $k>1$?

If that is the case, then since $k \ge 3$ we have $4^{b-a}+1\ge 9\cdot 4^{a-1}$. It follows that $2b-2a>2a-2+3$, and hence $b\ge 2a+1$.

But this is impossible. For note that since $1+4^a+4^b$ is a perfect square, we must have $$4^b+4^a+1 \ge (2^b+1)^2 =4^b+2^{b+1}+1$$ However, if $b \ge 2a+1$ then $4^{a}<2^{b+1}$, contradicting the above inequality.

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Ahh, I had started off similarly but could not even deal with when $2^{2b}+1$ is a perfect square so I was stuck. Thank you so much Andre, this seems to be precisely what is expected by my teacher. –  Bernhard Freeman Aug 12 '11 at 4:14
@anon: Have filled the gap. In that case $b$ turns out "too big" by the standard there is no square between $x^2$ and $(x+1)^2$. –  André Nicolas Aug 12 '11 at 5:02

According to Richard Guy (Unsolved Problems In Number Theory, Problem D10), Mignotte observed that the equation $n^2=2^a+2^b+1$ is completely solved by combining the results of Le Mao-Hua on $x^2=4q^n+4q^m+1$ with those of Tzanakis and Wolfskill. Le Mao-Hua wrote dozens of papers on Diophantine equations. I think the one Guy means is The Diophantine equation $x^2=4q^n+4q^m+1$, Proc Amer Math Soc 107 (1989) 599-604, MR 90b:11024. The T and W paper is The Diophantine equation $x^2=4q^{a/2}+4q+1$, J Number Theory 26 (1987) 96-116, MR 88g:11009.

EDIT: I think there is an elementary argument that works. I'll sketch it, in part.

As in Andre's (currently deleted) answer, we may assume $a\lt b$. I'll do the case where $b$ is odd, and leave you the case where $b$ is even. So, $b=2t-1$.

Show there aren't any squares between $4^b=(2^{2t-1})^2$ and $4^b+4^t+1$. Deduce that $a\ge t$.

From $4^b+4^a=(c+1)(c-1)$, deduce that $4^{a-(1/2)}$ divides $c+1$ or $c-1$, so $c\ge4^{a-(1/2)}-1$. Square both sides to get $4^b+4^a+1\ge4^{2a-1}-4^a+1$. Play around with the inequalities and deduce $a\le t$.

If you can fill in all the blanks, you're done.

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I have managed to locate some of those sources online and am finding them quite difficult to read, I think they are pitched above my level. I know that for this particular form of the problem, there is a completely elementary solution, as several of my high school classmates have solved this question already. I am very grateful for your time helping me. –  Bernhard Freeman Aug 12 '11 at 3:00
@Gerry Myerson: You and I were typing the same thing at the same time. And I am sure I have seen this before. –  André Nicolas Aug 12 '11 at 5:05