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Tahmid Hasan
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Unread post by Tahmid Hasan » Fri Jun 29, 2012 9:11 pm

Find all positive integers $x,y,z$ and $t$ such that $2^{x}3^{y}+5^{z}=7^{t}$
বড় ভালবাসি তোমায়,মা

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Phlembac Adib Hasan
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Re: JBMO-2012-4

Unread post by Phlembac Adib Hasan » Thu Aug 02, 2012 12:00 pm

This problem offers you two options: either do huge calculations or use Zsigmondy's theorem. Though none of them is attractive, but, at my point of view, the first option is much more honorable.
My Solution:
Taking mod $6$ shows $z$ is even.So let $z=2n$.Now we'll divide the problem into two cases:
Case:1 $x=1$
Taking mod $4$ and $5$ shows that $t$ must be odd and $y$ must be even.So\[2\cdot 3^y\equiv 4,1,2(mod\; 7)\]
\[5^z\equiv 4,2,1(mod\; 7)\]From the last two statements we can say \[2\cdot 3^y+5^z\equiv 1,2,3,4,5,6\not \equiv 0(mod\; 7)\]So a contradiction and no solution from here.
Case: 2 $x>1$
Taking mod $4$ gives $t$ is even.So let $t=2m$.And so the equation can be re-written as \[2^x3^y=(7^m+5^n)(7^m-5^n)\]
Now let $7^m+5^n=2^p3^q$ and $7^m-5^n=2^{x-p}3^{y-q}$.From the last two equation we find that
Notice that $2^{p-1}3^q$ and $2^{x-p-1}3^{y-q}$ must be co-prime so that $(i)$ and $(ii)$ remains valid.Hence $q=0$ or $y=q$ and $p=1$ or $x-p=1$.
Case: $2.1$ $q=0$
In this case $p=1$ is not possible because $7^m+5^n\ge 12>2=2^13^0=2^p3^q$.So $x-p=1$ and so $(i)$ and $(ii)$ convert into these equations:
By taking mod $3$, from $(iii)$ we can say $p-1$ is even, i.e. $x$ is even.(Because $x-p=1$)Also remember $p>1$.So $4|2^{p-1}$.
Now taking mod $3$ and mod $4$ on $(iv)$ gives $y$ is odd and $n$ is even.Let $p-1=2r$ and $n=2s$.So $(iv)$ can be re-written as $(2^r+5^s)(2^r-5^s)=3^y$.Notice that both $2^r+5^s$ and $2^r-5^s$ are powers of three.But their gcd must divide $2\cdot 5^s$.Then $2^r-5^s=3^0=1.......(v)$.If $r>1$, then a contradiction can be brought by taking mod $4$ on $(v)$.So $r=1$.But it will lead us to $(s,y)=(0,1)$which crosses the conditions.So no solution from here.
Case: $2.2$ $q=y$
Here also two cases:
Case: $2.2.1$ $p=1$
$(i)$ and $(ii)$ will convert into these equations:
We must have $x>2$ because $(vi)$ is not valid otherwise.Taking mod $3$ on $(vi)$ gives $x-2$ is even.So $4|2^{x-2}$.Let $x-2=2d$.
Now taking mod $4$ on $(vii)$ gives $y$ is also even.So let $y=2c$.Then $(vii)$ can be re-written as
\[5^n=(3^c-2^d)(3^c+2^d)........(viii)\]Notice that both $3^c-2^d$ and $3^c+2^d$ are powers of five.But their gcd must divide $2^{d+1}$.Then $3^c-2^d=1$.So we can write \[5^n=3^c+2^d=(2^d+1)+2^d=2^{d+1}+1\]\[\Longrightarrow 2^{d+1}=5^n-1\] Notice that \[5^{2k}-1\equiv (-1)^{2k}-1\equiv 0(mod\; 6)\]
But $3\not |2^{d+1}$.So n is odd.So let $n=2u+1$.Then \[5^n-1\equiv 5^{2u}5-1\equiv 5-1=4(mod\; 8)\]As $d>0$, So $d+1=2$ and $d=1$.Then $n=c=1$.So $y=2,x=4$.But $(vi)$ is not satisfied with these values of $x$ and $y$.So again no solution.
Case: $2.2.2$ $x-p=1$
$(i)$ and $(ii)$ will convert into these equations:
From our previous case we know $p>1$.But if $p>2$, then by taking mod $4$ we can bring a contradiction to $(x)$.So $p=2$ and $x=3$.
So $(x)$ can be converted to\[5^n=2\cdot 3^y-1.......(xi)\]
It follows from LTE that $3^{y-1}|n$.But then \[5^n\ge 5^{3^{y-1}}\ge 2\cdot 3^y-1\]with equality iff $y=1$.(The last inequality can be brought by simple induction or calculus so this part is skipped)Then $n=m=1$.So $t=z=2$.
So only solution is $(x,y,z,t)=(3,1,2,2)$ which is a correct solution.
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