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How many integral solutions of $a,\ b,\ c$ are there such that $$2^a \cdot 3^b + 9 = c^2.$$

we can get that $$2^a \cdot3^b = (c-3)(c+3) $$

we can make cases if $b \ge 2$ then $c=3k$

then $$2^a \cdot 3^{b-2} = (k-1)(k+1) $$

Now if $a \ge 2$ then $ k = 2l+1$

$2^{a-2} \cdot 3^{b-2} = l \cdot(l+1) $

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At this point, we see that $2$ can only divide one of the terms on the right; same with $3$. What powers of $2$ are exactly one less (or greater than) a power of $3$? – Mike Miller Nov 27 '13 at 16:44
i have got 5 solutions..values of c are 5, 6 ,15, 51, 21,19. – maths lover Nov 27 '13 at 16:46
My mistake - those are the solutions. – Mike Miller Nov 27 '13 at 16:53
@mathslover I presume you meant $9$ instead of $19$ (and yes, you listed six solutions in total :-) ) – Peter Košinár Nov 27 '13 at 17:09
but how to get these.. i got it from trial error.. – maths lover Nov 27 '13 at 17:48

1 Answer 1

up vote 1 down vote accepted

Well, standard techniques work, albeit a bit tedious to type out.

There are twelve solutions:

$$(a, b, c)=(4, 0, \pm 5), (0, 3, \pm 6), (3, 2, \pm 9), (3, 3, \pm 15), (5, 4, \pm 51), (4, 3, \pm 21)$$

Note that $2^a3^b=c^2-9 \in \mathbb{Z}$ so $a \geq 0, b \geq 0$. WLOG assume $c \geq 0$ (if $c<0$ replace $c$ by $-c$)

If $b=0$, then


We must thus have

$$c+3=2^{\alpha}, c-3=2^{\beta}, a=\alpha+\beta, \alpha \geq \beta \geq 0$$

Now $2^{\beta} \| (2^{\alpha}-2^{\beta})=6$, so $\beta=1$, so we get $\alpha=3$. This gives $a=\alpha+\beta=4$ and $c=2^{\beta}+3=5$. We get the solution $(a, b, c)=(4, 0, 5)$.

Otherwise $b \geq 1$. Then $\pmod{3}$ gives $3 \mid c$, so that $9 \mid c^2$ and we get $b \geq 2$. Put $c=3k$ and $b-2=b' \geq 0$. Then we have


Now if $a=0$, then


We must thus have

$$k+1=3^{\gamma}, k-1=3^{\delta}, b'=\gamma+\delta, \gamma \geq \delta \geq 0$$

Now $3^{\delta} \|(3^{\gamma}-3^{\beta})=2$, so $\delta=0$ and we get $\gamma=1$. This gives $b'=\gamma+\delta=1, b=b'+2=3, k=3^{\delta}+1=2, c=3k=6$. We get the solution $(a, b, c)=(0, 3, 6)$.

Otherwise $a \geq 1$. Then $\pmod{2}$ gives $k$ odd, so $4 \mid (k-1)(k+1)$ so $a \geq 2$. Write $k=2l+1$ and $a-2=a' \geq 0$. Clearly $k>1$ so $l \geq 1$. We get


Since $(l, l+1)=1$ this implies that we have

$$\begin{cases} l=1, l+1=2^{a'}3^{b'} \\ l=2^{a'}, l+1=3^{b'} \\ l=3^{b'}, l+1=2^{a'} \\ l=2^{a'}3^{b'}, l+1=1 \end{cases}$$

The first case gives $2=l+1=2^{a'}3^{b'}$ so $a'=1, b'=0, a=a'+2=3, b=b'+2=2, c=3k=3(2l+1)=9$, giving the solution $(a, b, c)=(3, 2, 9)$.

The last case is clearly impossible, for $l \geq 1$.

The second case gives $3^{b'}-2^{a'}=(l+1)-l=1$, and the third case gives $2^{a'}-3^{b'}=(l+1)-l=1$.

Note: If we were lazy, at this point we could overkill using Mihailescu's theorem (used to be called Catalan's conjecture) and conclude that apart from $3^2-2^3=1$, it suffices to check the cases $a'=0, 1$ and $b'=0, 1$.

In the spirit of contest math, let us provide an elementary approach, and not overkill. Unfortunately this means more things to type.

Let us start with the second case:

$$3^{b'}=l+1, 2^{a'}=l, 3^{b'}-2^{a'}=1$$

Clearly $b' \not =0$, so $b' \geq 1$. Also $a'=0$ does not give a solution, so $a' \geq 1$. If $a'=1$, we get $b'=1$ and $l=2$. This gives $a=b=3$ and $c=3k=3(2l+1)=15$, so we get the solution $(a, b, c)=(3, 3, 15)$.

Otherwise $a' \geq 2$. Taking $\pmod{4}$ gives $2 \mid b'$. Write $b'=2d$ then we get


We get

$$3^d+1=2^w, 3^d-1=2^x, a'=w+x, w \geq x \geq 0$$

Then $2^x \|(2^w-2^x)=2$, so $x=1$ and $w=2$. We get

$$d=1, b'=2d=2, b=b'+2=4, a'=w+x=3, a=a'+2=5$$ $$l=2^{a'}=8, c=3(2l+1)=51$$

We get the solution $(a, b, c)=(5, 4, 51)$.

Finally let us look at the third case:

$$l=3^{b'}, l+1=2^{a'}, 2^{a'}-3^{b'}=1$$

If $b'=0$, we get $a'=1$ and $a=3, b=2, l=3^{b'}=1, c=3(2l+1)=9$. We get the solution $(a, b, c)=(3, 2, 9)$, which we already got. (Overlap with first case)

Otherwise $b' \geq 1$. Taking $\pmod{3}$ gives $2 \mid a'$. Writing $a'=2e$ gives


We thus have

$$2^e+1=3^y, 2^e-1=3^z, b'=y+z, y \geq z \geq 0$$

Now $3^z \| 3^y-3^z=2$, so $z=0$ and we get $y=1$. This gives

$$e=1, a'=2e=2, b'=y+z=1, a=a'+2=4, b=b'+2=3$$ $$l=3^{b'}=3, c=3(2l+1)=21$$

We get the solution $(a, b, c)=(4, 3, 21)$.

To conclude, the solutions with $c \geq 0$ are

$$(a, b, c)=(4, 0, 5), (0, 3, 6), (3, 2, 9), (3, 3, 15), (5, 4, 51), (4, 3, 21)$$

Therefore all solutions are given by

$$(a, b, c)=(4, 0, \pm 5), (0, 3, \pm 6), (3, 2, \pm 9), (3, 3, \pm 15), (5, 4, \pm 51), (4, 3, \pm 21)$$

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