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Source: Romania JBMO TST 2015 Day 1 Problem 4

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#1 h May 14, 2015, 6:52 AM • 2 Y

Y by ahmedosama, Adventure10

Solve in nonnegative integers the following equation :
$21^x+4^y=z^2$

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BSJL

#2 h May 14, 2015, 7:10 AM • 2 Y

Y by Adventure10, Mango247

Rewrite the condition:

$3^x \cdot 7^x=(z-2^y)(z+2^y)$

Since $z-2^y, z+2^y$ can't be the multiples of 3,7 at the some time, so we have two cases.

Case: $\boxed{z-2^y=1,z+2^y=21^x}$

We have $2^{y+1}=21^x-1$ . Notice that $5|(21^x-1)$ , thus, there isn't any solution.

Case: $\boxed{z-2^y=3^x,z+2^y=7^x}$

In this case, we have $2^{y+1}=7^x-3^x$ , which can be solve immediately by Zsigmody.

The only solution is $x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$

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#3 h May 14, 2015, 9:38 AM • 1 Y

Y by Adventure10

BSJL wrote:

In this case, we have $2^{y+1}=7^x-3^x$ , which can be solve immediately by Zsigmody.

The only solution is $x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$

By Zsigmondy, more generally $a^n+b^n$ has at least two prime divisors when $a>b>0, n\ge 2, (a,b,n)\neq (2,1,3)$ .

$a^n - b^n$ has at least two prime divisors when $a>b+1>1, n\ge 2, (a,b)=1, (a,b,n)\neq (a, 2^m-a, 2)$ for any integer $m$ .

Here $7>3+1>1, (7,3)=1, (a,b,n)\neq (a,2^m-a,2)$ for any integer $m$ , since $x\neq 2$ by checking.

$x\ge 2$ gives contradiction by above Zsigmondy lemma, so $x\in\{0,1\}$ . $x\neq 0$ . $x=1$ gives $(x,y,z)=(1,1,5)$ .

Don't confuse set notation. It is false that $\{x,y,z\}=\{1,1,5\}$ is a solution (clearly $z\neq 1$ ).

Zsigmondy is overkill here (the proof of Zs isn't elementary). $x$ even gives $2^{y+1}=(7^{x/2}-3^{x/2})(7^{x/2}+3^{x/2})$ .

$7^{x/2}+3^{x/2}\equiv 2\pmod {4}$ , so $7^{x/2}+3^{x/2}=2$ and $x=0$ , contr.

So $x$ is odd. mod $8$ gives $7^x-3^x\equiv 4\pmod {8}$ , so $y=1$ , and $(x,y,z)=(1,1,5)$ .

This post has been edited 7 times. Last edited by randomasdf97, May 14, 2015, 8:09 PM

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#4 h Jun 1, 2015, 2:31 PM • 2 Y

Y by Adventure10, Mango247

ComplexPhi wrote:

Solve in nonnegative integers the following equation :
$21^x+4^y=z^2$

If $x = 0$ we get $(z - 1)(z + 1) = 4^y$ ,if we look at equation $modulo$ $4$ we obtain,if $z-1 \equiv 0($ $mod$ $4)$
$\implies$ $z+1 \equiv 0$ $(mod$ $4)$ ,and similarly if $z - 1 \equiv 0$ $(mod$ $4)$ we obtain contradiction so $x \neq 0$ .
If $y = 0$ we get $(z - 1)(z + 1) = 21^x$ , $z-1\neq 1$ and $z+1\neq 1$ .
If both $z-1$ , $z+1$ are divisible with 21 we obtain that $gcd(z-1,z+1)\leq 2$ ,contradition.
If $z - 1 = 3^x$ , $z+1 = 7^x$ $\implies$ $7^x - 3^x = 2$ ,contradiction $7^x - 3^x$ is divisible with 4.
If $z - 1 = 7^x$ , $z+1 = 3^x$ $\implies$ $7^x < 3^x$ ,contradiction.
We obtain that $y\neq 0$ .
$3^x\cdot7^x = (z - 2^y)(z+2^y)$ .
If $z - 2^y = 1$ $\implies$ $z = 2^y + 1$ and $4^y+2^{y+1}+1=4^y+21^x$ or $21^x-1^x = 2^{y+1}$ ,but $21^x - 1^x$ if divisible with $20$ ,contradiction.
If $z +2^y = 1$ $\implies$ $y = 0$ ,contradiction.
If both $z - 2^y$ and $z + 2^y$ are divisible with 21 $\implies$ $21\equiv 0$ $(mod$ $2)$ ,contradiction.
If $z - 2^y = 3^x$ and $z + 2^y = 7^x$ $\implies$ $3^x + 2^{y+1} = 7^x$ .
If $y\ge 2$ $\implies$ $3^x\equiv 0$ $(mod$ $8)$ but $3^x\equiv 3,1$ $(mod$ $8)$ and $7^x\equiv -1,1$ $(mod$ $8)$ ,contradiction $\implies$ $y = 1$ , $x = 1$ and $z = 5$ .
If $z - 2^y = 7^x$ and $z + 2^y = 3^x$ $\implies$ $3^x > 7^x$ ,contradiction.
So the single solution is $(x,y,z) = (1,1,5)$ .

This post has been edited 6 times. Last edited by neverlose, Feb 8, 2016, 4:43 PM

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#5 h Sep 25, 2023, 2:46 PM

Y by

To fill in, here is an elementary way to solve $2^{y+1} = 7^x - 3^x$ for $y\geq 2$ . By mod $8$ we get that $x$ is even and so $2^{y+1} = (7^{x/2} - 3^{x/2})(7^{x/2} + 3^{x/2})$ , thus (as the two factors have gcd $2$ ) $7^{x/2} - 3^{x/2} = 2$ , which is impossible since $3^{x/2} \geq 9$ and $(7/3)^{x/2} - 1 > 5$ for $x\geq 4$ .

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#11 h Apr 22, 2025, 6:58 AM • 1 Y

Y by Turkish_sniper

If $y=0 \implies 3^x.7^x=(z-1)(z+1)$
By Euclidean Algorithm, there is 2 cases.
Case 1: $z+1=7^x$
$z-1=3^x$
So $7^x-3^x=2$ which is impossible.
Case 2 $z-1=1$
$z+1=21^x$ is a contradiction.

If $y=1 \implies 3^x.7^x=(z-2).(z+2)$
By Euclidean Algorithm, again there is 2 cases.
Case 1: $z-2=3^x$
$z+2=7^x$
$\implies 7^x-3^x=4$
Clearly there is only 1 solution. $\boxed{(x,y,z)=(1,1,5)}$
Case 2: $z-2=1$
$z+2=21^x$ is impossible.

If $y>1$
By $mod 8$ x is even.
So the new equation is $2^{2y}=(z-21^{x/2}).(z+21^{x/2})$
By Euclidean Algorithm $z-21^{x/2}=2$ $z+21^{x/2}=2^{2y-1}$
$\implies 2.21^{x/2}=2^{2y-1}-2$
If we divide both sides by 2 $21^{x/2}=2^{2y-2}-1$ And this is also a contradiction by $mod 8$
So the only solution is $\boxed{(x,y,z)=(1,1,5)}$

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#12 h Apr 22, 2025, 1:52 PM

Y by

The answer is $\boxed{(1,1, 5)}$ only. We have that $z$ is odd, so $(z^{2}-2^{y})(z^{2}+2^{y}) = 3^{x}*7^{x},$ and thus $2^{y+1} = 7^{x}-3^{x}.$ Now, $\nu_{2}(7^{x}-3^{x}) = 3+\nu_{2}(x)-1 = \nu_{2}(x)+2,$ so $2^{\nu_{2}(x)+2} = 7^{x}-3^{x} \le 4x,$ which is true only when $x \le 1,$ so our solution is $x=1$ , $y=1$ , and then $z=5.$

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