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Find all integers $x,y,z$ such that $5^x+7^y=2^z$.

This one comes from an online contest that I arranged some years ago, and I can assure that a completely elementary solution exists.

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So you know the answer to your question? Hey, that's not fair :) –  Lord Soth Apr 1 '13 at 23:36
    
$\,(2,1,5)\,,\,(2,2,6)\,$ ... –  DonAntonio Apr 1 '13 at 23:36
    
$(2,2,6)$ is not a solution :O –  Paolo Leonetti Apr 1 '13 at 23:41
    
Yeah, it is not: it's a pity $\,2^6\neq 74\,$....tsk,tsk,tsk. Well, at least the first one is. :) –  DonAntonio Apr 1 '13 at 23:44
    
That happens :P –  Paolo Leonetti Apr 1 '13 at 23:49
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3 Answers 3

A "completely elementary solution":

We shall use only modular arithmetic.

Since $\gcd(2, 5, 7)=1$, $x, y, z$ are non-negative.

If $x=0$, then $1+7^y=2^z$. If $z \geq 4$, taking $\pmod{16}$ gives $1+7^y \equiv 0 \pmod{16}$, a contradiction. Thus $z \leq 3$, and checking gives the solutions $(x, y, z)=(0,0, 1), (0, 1, 3)$.

Otherwise $x \geq 1$, so $2^z=5^x+7^y \geq 5+1$, so $z \geq 2$. Taking $\pmod{4}$ gives $1+(-1)^y \equiv 0 \pmod{4}$, so $2 \nmid y$, implying that $y \geq 1$. We have $2^z=5^x+7^y \geq 12$, so $z \geq 4$.

Taking $\pmod{16}$, we get $5^x+7^y \equiv 0 \pmod{16}$, so $x \equiv 2 \pmod{4}, 2 \nmid y$.

Taking $\pmod{5}$, we get $2^y \equiv 7^y \equiv 2^z \pmod{5}$, so $y \equiv z \pmod{4}$.

If $y \equiv z \equiv 3 \pmod{4}$, then taking $\pmod{13}$, we get $-1+7^y \equiv 5^x+7^y \equiv 2^z \pmod{13}$. We have $7^y \equiv 5, 6, 2 \pmod{13}, 2^z \equiv 8, 11, 7 \pmod{13}$, which is impossible.

Thus $y \equiv z \equiv 1 \pmod{4}$. Again taking $\pmod{13}$, we get $-1+7^y \equiv 5^x+7^y \equiv 2^z \pmod{13}$. We have $7^y \equiv 7, 11, 8 \pmod{13}, 2^z \equiv 2, 6, 5 \pmod{13}$, so $y \equiv 1 \pmod{3}, z \equiv 2 \pmod{3}$.

Taking $\pmod{7}$, we get $2^x \equiv (-2)^x \equiv 5^x \equiv 2^z \pmod{7}$, so $x \equiv z \equiv 2 \pmod{3}$.

Combining the information so far, we have $(x, y, z) \equiv (2, 1, 5) \pmod{12}$. For convenience, let us write $x=12a+2, y=12b+1, z=12c+5, a, b, c \geq 0$, so the equation becomes:

$$5^{12a+2}+7^{12b+1}=2^{12c+5}$$

Now so far everything we have done still allows for the solution $(2, 1, 5)$.

Let us consider $b \geq 1$ first. Then taking $\pmod{49}$, we get $2^{-(6a+1)} \equiv 25^{6a+1} \equiv 5^{12a+2} \equiv 2^{12c+5} \pmod{49}$, so $15^{2c+a+1} \equiv 2^{12c+6a+6} \equiv 1 \pmod{49}$. Thus $7 \mid 2c+a+1$.

Now let's take $\pmod{43}$. We shall get a contradiction. We have $5^{12} \equiv -2 \pmod{43}, 7^{12} \equiv 1 \pmod{43}, 2^{12} \equiv 11 \pmod{43}$. Thus $25(-2)^a+7 \equiv 32(11)^c \equiv 32(-2)^{5c} \equiv 32(-2)^{a+1} \equiv (-64)(-2)^a \pmod{43}$. This gives: $$89(-2)^a \equiv -7 \pmod{43}$$ $$3(-2)^a \equiv 36 \pmod{43}$$ $$(-2)^a \equiv 12 \pmod{43}$$

But $(-2)^a \equiv 41, 4, 35, 16, 11, 21, 1 \pmod{43}$, so we get a contradiction.

Therefore $b=0$, and we get:

$$5^{12a+2}+7=2^{12c+5}$$

By now I am really tempted to just say "This is a special case of the Ramanujan square equation, for which solutions are known." Nevermind. Let's continue.

If $c=0$, then $a=0$, and we get the solution $(x, y, z)=(2, 1, 5)$.

Otherwise $c \geq 1$. Taking $\pmod{64}$, we get $25(17^a)+7 \equiv 5^{12a+2}+7 \equiv 0 \pmod{64}$. Thus $a \equiv 2 \pmod{4}$. Put $a=4d+2$, so we get

$$5^{48d+26}+7=2^{12c+5}$$

Taking $\pmod{17}$, we get $16 \equiv 5^{10}+7 \equiv 5^{48d+26}+7 \equiv 2^{12c+5} \equiv 32(-1)^c \pmod{17}$, giving a contradiction.

To conclude, the only solutions are $(x, y, z)=(0, 0, 1), (0, 1, 3), (2, 1, 5)$.

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That is almost the same of mine, well done: +1. –  Paolo Leonetti Apr 3 '13 at 16:20
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I'm going to post a possible proof and maybe with some important mistakes,please check it and let me see where is wrong.

(1)$x,y,z>0$. Since $Z(\sqrt{-7})$ is an Euclidean domain,the factorization is unique,and the elements are those $z=a+bt$, and $N(z)=a^2+ab+2b^2$,where $t=\frac{1+\sqrt{-7}}2$.The units are $z=±1$,and $2=t(1-t)$.

$5^x+7^y=2^z$,if $z>2$ then $$5^x+7^y\equiv 1^x+(-1)^y\equiv 0\pmod4$$ hence $y\equiv 1\pmod2$. and $$5^x+7^y\equiv 5^x+(-1)^y\equiv 5^x-1\equiv 0\pmod8$$ hence $x\equiv 0\pmod2$. $$5^x+7^y-2^z\equiv(-2)^x+0-2^z\equiv2^x-2^z\equiv 0\pmod7$$ hence $x\equiv z\pmod3$. $$5^x+7^y-2^z \equiv 0+2^y-2^z \equiv 0\pmod5$$ hence $y\equiv z\pmod 4$.

We only need $x\equiv 0\pmod2,y\equiv 1\pmod2$. Let $u=5^{x/2},v=7^{(y-1)/2},w=z-2$, here $w>0$,and $u^2+7v^2=2^{w+2}$. We get $$(u+v(2t-1))*(u-v(2t-1))=2^{w+2}$$ $$(\frac{u-v}2+vt)*(\frac{u+v}2-vt)=2^w=t^w*(1-t)^w$$ here $t$ and $1-t$ are both prime in $Z(\sqrt{-7})$.

Since $\frac{u-v}2+vt=\frac{5^{x/2}-7^{(y-1)/2}}2+7^{(y-1)/2}t$, and $7^{(y-1)/2}$ is odd, so $GCD(\frac{u-v}2+vt,\frac{u+v}2-vt)=1$ hence $$\frac{u-v}2+vt=t^w,\frac{u+v}2-vt=(1-t)^w$$ or change the order(we can replace $v$ with $-v$,so it doesn't matter). Hence $$u=t^w+(1-t)^w,v=\frac{t^w-(1-t)^w}{2t-1)}$$ as $y>1$ , $v>1$ and $u\equiv0 \pmod 5,v\equiv0 \pmod 7$.

With a little compute,we get $w\equiv 21 \pmod{42}$.However,it leads to $u\equiv0 \pmod 13$,which is a contradiction. So $y$ cannot great than $1$.

If $y=1$ then $5^x+7=2^z$,which is Ramanujan equation:$u^2+7=2^z$,so $x=2,z=5$.

(2)If $x=0$,then $1+7^y=2^z$.if $z>3$,then $1+7^y\equiv 0\pmod{16}$,that's impossible.

In the case $z<=3$,we have $(x,y,z)=(0,0,1)(0,1,3)$.

If $x>0$,as we have proved in (1),$y$ is odd, hence $y>0$,and $z>0$.

So $(x,y,z)=(0,0,1)(0,1,3)(2,1,5)$.

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Thumb up. You got the non-elementary solution that I did know, but really in a few. Just a thing to avoid a reference: you can skip the Ramanujan equation, just looking in sequence at modules $25$, $32$ and $17$. Just to complete, at this point if $\min\{x,y,z\}<0$ then also $\max\{x,y,z\}<0$, and getting immediately a contradiction. –  Paolo Leonetti Apr 2 '13 at 11:10
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I could partially solve this.

$5^x + 7^y = 2^z$

Firstly, we will write down the last digits of the powers of $5,7,2$ respectively.

For any power of '$5$' the last digit is always $'5'$.

$5, 5, 5, 5$

For powers of '$7$' the last digits are as follows and they repeat in cycles of $4$:

$7, 9, 3, 1$

For powers of '$2$' the last digits are as follows and they repeat in cycles of $4$:

$2, 4, 8, 6$

So By adding the term ending in '$5$' to that ending in '$7$' will give '$2$' and this is the case for the other $3$ terms.

Hence, if y = $(4m+1)$ , then z = $(4n+1)$.

Therefore, the difference between $y$ and $z$ is always a multiple of $4$. Also one must keep in mind that $2^z - 7^y$ should always be positive because $5^x$ can never be negative.

So $z=5,y=1$ satisfies this and this leads to the solution $x=2,y=1,z=5$ or $(2,1,5)$.

There clearly exist no other solution for higher values of '$z$', but i am unable to prove it.

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I have taken the liberty of very slightly editing this because I found the terminology confusing at first. –  Potato Apr 2 '13 at 5:16
    
If you remove the "clearly" the problem begins ;) To sum up, lsp just proved that: "If $(x,y,z)$ is a solution then $4\mid z-y$". –  Paolo Leonetti Apr 2 '13 at 7:17
    
@PaoloLeonetti: But $(0,0,1)$ and $(0,1,3)$ violate that. –  Ross Millikan Apr 2 '13 at 15:36
    
Indeed it was assumed implicitely that $x\ge 1$; in rough terms, half problem is missed, i.e. all cases $x\le 0$; except that, we are talking about trivial cases.. I am still waiting if someone finds a elementary solution different from mine –  Paolo Leonetti Apr 2 '13 at 17:29
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@PaoloLeonetti Why don't you go ahead and post the elementary solution you do know, then? –  Potato Apr 3 '13 at 1:08
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