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here is another problem I did not manage to solve in the contest I mentioned in my previous question.

Determine the number of integer solutions $(i, j)$ of the equation: $3i^2 + 2j^2 = 77 \cdot 6^{2012}$.

Applying logarithms is not useful, since on the left hand side we have a sum; I also tried some algebraic manipulations that led me to nothing useful. Is there a simple solution to the problem?

Thank you,

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up vote -2 down vote accepted

$A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$

$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :

$i= \pm 5x \Rightarrow i = \pm 5 \cdot 6^{1006}$

$j= \pm x \Rightarrow j = \pm 6^{1006}$

$R_A :$

$(i,j) \in \{(-5 \cdot 6^{1006},-6^{1006}),(-5 \cdot 6^{1006},6^{1006}),(5 \cdot 6^{1006},-6^{1006}),(5 \cdot 6^{1006},6^{1006})\}$

$B)~ i^2=9x^2 ~\text {and}~ j^2=25x^2$

$3 \cdot 9x^2+2 \cdot 25x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :

$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$

$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$

$R_B :$

$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$

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Wow, this is brilliant! Simple and neat! I am choosing your answer as best answer because in my opinion it's the simplest and intuitive one. – rubik Mar 11 '12 at 10:51
I don't understand. So what if the ratio $i/j$ is neither 5 nor 5/3? – Zsbán Ambrus Mar 11 '12 at 11:10
I'm with @ZsbánAmbrus I can't follow this at all. It seems completely unintuitive. – Pureferret Mar 11 '12 at 13:56
@ZsbánAmbrus: Can you elaborate a bit more? – rubik Mar 12 '12 at 13:33

Observe that $i$ has to be even, so do $j$ because $3i^2$ and the right side can be divided by 4. Let $i'=2i$, $j'=2j$; the equation becomes $3i'^2+2j'2=77 \cdot 3^{2012} \cdot 2^{2010}$. The same argument applies 1006 times, and we get something like $3i^2+2j^2=77 \cdot 3^{2012}$. Now do the same looking the divisibility by 3. It works. Finaly check $3i^2+2j^2=77$

Edit: It's my first posting here, so hello to all from France !

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Welcome to math.stackexchange ama! – Ragib Zaman Mar 11 '12 at 9:49

Considering the equation modulo $3$, we must have $2j^2 \equiv 0 $ which implies that $ j=3k$ for some $k\in \mathbb{N}$.

$ 3i^2 + 18k^2 = 77 \cdot 6^{2012} $ so $$ i^2 + 6k^2 = 77 \cdot 2^{2012} \cdot 3^{2011}.$$

which now implies $i=6l $ for some $l\in \mathbb{N}.$ Thus

$$ 6l^2 + k^2 = 77 \cdot 2^{2011} \cdot 3^{2010}.$$

Similarly, $k=6m$ for some $m\in\mathbb{N} $ and

$$ l^2 + 6m^2 = 77 \cdot 2^{2010} \cdot 3^{2009}.$$

A pattern is emerging that we will keep on having this divisibility by 6 until we reach some $a$ and $b$ such that $$ a^2 + 6b^2 = 77 \cdot 2 =154 $$

which (by checking the small number of possible cases) has solutions $ (a,b) = (10,3)$ and $ (2,5)$ in the naturals, or by varying all possible sign combinations, there are 8 solutions in the integers. Now to find the solutions of the original equation, simply back substitute the factors of 6 we kept extracting. For this problem however that is not required since only the number of solutions is asked, and the answer is 8.

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+1: I like your method because it's intuitive and very understandable. – rubik Mar 11 '12 at 10:56

The righthand side is even, as is $2j^2$, so $i$ must be even. Say $i=2^nm$, where $m$ is odd and $n\ge 1$, and $j=2^k\ell$, where $\ell$ is odd and $k\ge 0$; then $3m^22^{2n}+\ell^2 2^{2k+1}=77\cdot6^{2012}$. Let $a=\min\{2n,2k+1,2012\}$; then $3m^22^{2n-a}+\ell^2 2^{2k+1-a}=77\cdot3^{2012}2^{2012-a}$, where the exponents on the $2$’s are all non-negative, and at least one is $0$. Clearly it’s not possible for exactly two of these exponents to be positive, and since the first two have opposite parity, it’s not possible for all three to be zero, so exactly one is positive and two are zero. The first and third have the same parity, so we must have $2n-a=2012-a=0$ and $2k+1-a>0$. In particular, $a=2012$, and $n=1006$.

To recapitulate, at this point we know that $i=2^{1006}m$ for some odd $m$, and $j=2^k\ell$ for some odd $\ell$ and $k\ge 1006$, and we can let $k\,'=k-1006$ and divide through by $2^{2012}$ to get $$3m^2+2^{2k\,'+1}\ell^2=77\cdot3^{2012}\;.\tag{1}$$

Now perform a similar analysis considering factors of $3$ instead of $2$. If $3^b$ and $3^c$ are the highest powers of $3$ dividing $m$ and $\ell$, respectively, then $3m^2$ has $2b+1$ factors of $3$, $2^{2k\,'+1}\ell^2$ has $2c$, and the righthand side has $2012$. Reasoning as before, we see that this is possible only if $2c=2012$ and $2b+1>2012$, so that $c=1006$ and $b\ge 1006$. Thus, if we set $b\,'=b-1006$ and divide through by $3^{2012}$, we can further reduce $(1)$ to


where $m=3^{2b+1}u$, $\ell=3^{1006}v$, and $u$ and $v$ are not divisible by $2$ or $3$.

Plainly $0\le b\,'\le 1$ and $0\le k\,'\le 2$, so there are at most six solutions. Moreover, the only possible values for $u$ and $v$ are $1$ and $5$: $7$ is already too large.

  • If $u=v=1$, the lefthand side is either too large or at most $3^3+2^5=59$, so we get no solutions.

  • If $u=5$, $b\,'$ must be $0$, and we get the solution $3\cdot25+2\cdot1=77$.

  • If $v=5$, $k\,'$ must be $0$, and we get the solution $3^3\cdot1+2\cdot25=77$.

The original equation therefore has two solutions in positive integers and therefore eight in integers (since we can change the sign of $i$ or $j$ independently in each solution).

This can probably be done more elegantly; I wrote it up as I was working it out.

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Doesn't 2 solutions in the naturals imply 2*4 = 8 solutions in the integers, since for each (+,+) we also have (-,+), (+,-) and (-,-). – Ragib Zaman Mar 11 '12 at 9:51
@Ragib: Yep; momentary slip of the mind. Thanks. – Brian M. Scott Mar 11 '12 at 9:56
+1: Thank you; your method is a bit more complicated than the other ones but you showed how to reason. Probably your last lines served as basis for pedja's answer (I'm referring to $3\cdot25+2\cdot1=77$ and $3^3\cdot1+2\cdot25=77$). – rubik Mar 11 '12 at 11:11

I find it easiest to proceed as follows. First note that $j$ must be divisible by $3$ and $i$ must be even. But then $3i^2$ is divisible by $9,$ so $i$ is divisible by $3,$ and similarly $j$ is even. Suppose then that $6^k$ divides ${\rm gcd}(i,j)$ but $6^{k+1}$ does not. Then $ 1 \leq k \leq 1006.$ Write $i = 6^{k}s$ and $j = 6^k t.$ If $k < 1006,$ we have $36$ divides $3s^2 + 2t^2,$ and again $3$ divides both $s$ and $t$, and $s$ and $t$ are both even, contadicting the choice of $k,$ since $6$ then divides ${\rm gcd}(s,t).$ Hence we are reduced to finding the number of integer solutions to $3s^2 + 2t^2 = 77.$ Now $s$ must be odd, so $3s^2 \equiv 3$ (mod $8$) and hence $2t^2 \equiv 2$ (mod $8$), and $t$ is also odd. Thus $s$ an $t$ are both odd. Clearly $|s| \leq 5.$ If $|s| = 5,$ then $|t| = 1,$ while if $|s| = 3$ then $|t| = 5.$ If $|s| = 1,$ then there is no solution for $t.$ Hence the solutions for $s$ and $t$ are $(5,1),(5,-1),-5,1),(-5,-1), (3,5),(3,-5),(-3,5),(-3,-5),$ and each solution for $(s,t)$ yields the unique solution $6^{1006}(s,t)$ for $(i,j).$

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Hint $\ \ \ $ Prime $\rm\:p\nmid a,\:\ p^2\ |\ n = a\: x^2 + p\:y^2\:\Rightarrow\: p\ |\ x^2\:\Rightarrow\: p\ |\ x\:\Rightarrow\: p\ |\ y^2\:\Rightarrow\: p\ |\ y $

Therefore, we deduce $\rm\:\ n/p^2 = a\:\bar x^2 + p\:\bar y^2,\ $ for $\rm\ \bar x = x/p,\ \bar y = y/p \in\mathbb Z$

So for $\rm\: p = 3,2,\ \ \ 3\:i^2 + 2\:j^2 = 77 \cdot 6^{2012}\:\Rightarrow\: 3\:\bar i^2 + 2\:\bar j^2 = 77\ $ for $\rm \bar i = i/6^{1006},\:\bar j = j/6^{1006}\in\mathbb Z$

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