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Lifting the Exponent

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Source: Japan TST 2024 p6

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#1 h Oct 5, 2024, 2:25 AM • 1 Y

Y by MathLuis

Find all quadruples $(a, b, c, d)$ of positive integers such that
$2^a3^b + 4^c5^d = 2^b3^a + 4^d5^c.$

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#2 h Apr 4, 2025, 10:24 AM • 3 Y

Y by alexanderhamilton124, L13832, EeEeRUT

Suppose that $a=b$ , then $4^c5^d=4^d5^c$ . It follows that $c=d$ . Similarly when $c=d$ , we have $a=b$ .
Notice that the equation is equivalent to
$2^a3^b-2^b3^a=4^d5^c-4^c5^d\implies 2^a3^a\left(3^{b-a}-2^{b-a}\right)=4^d5^d\left(5^{c-d}-4^{c-d}\right).$ If $a<b$ and $c<d$ : $3^{b-a}-2^{b-a}>0$ , but $5^{c-d}-4^{c-d}<0$ . This leads to a contradiction.
If $a>b$ and $c>d$ : $3^{b-a}-2^{b-a}<0$ , but $5^{c-d}-4^{c-d}>0$ . This leads to a contradiction.
So that $a<b$ and $c>d$ or $a>b$ and $c<d$ . We may assume without loss of generality that $a<b$ and $c>d$ (the other case can be handled similarly).
First, considering $\nu_2$ both sides gives $a=2d$ . So, our equation becomes
$3^{2d}\left(3^{b-2d}-2^{b-2d}\right)=5^d\left(5^{c-d}-4^{c-d}\right).$ Since $d$ is a positive integer, $3\mid 5^{c-d}-4^{c-d}$ . Which implies that $2\mid c-d$ . Analogously, $5\mid 3^{b-2d}-2^{b-2d}$ . Which implies $2\mid b-2d\implies2\mid b$ .
We may write $b=2b'$ for some positive integer $b'$ , and check $\nu_5$ from both sides gives:
$\nu_{5}\left(3^{2b'-2d}-2^{2b'-2d}\right)=1+\nu_5(b'-d)=d$ where the above equation is regarded to the lifting-the-exponent lemma. Also, by checking $\nu_3$ :
$\nu_3\left(5^{c-d}-4^{c-d}\right)=2+\nu_3(c-d)=2d.$ It yields that $5^{d-1}\mid b-2d$ and $3^{2d-2}\mid c-d$ .
It's natural to take mod $8$ , and get $1-2^{b-2d}\equiv 5^d\pmod{8}$ . We may split into two cases:

Case 1: If $d$ is an odd integer, then we have $5^d\equiv 5\pmod{8}$ , and $2^{b-2d}\equiv 4\pmod{8}$ .
It is enough to conclude that $b-2d=2$ . But, we have $5^{d-1}\mid b-2d$ . That is $d=1$ and $b=4$ .
Plugging back into above equation gives
$9\left(3^{2}-2^2\right)=5\left(5^{c-1}-4^{c-1}\right)\implies c=3.$ Therefore, $(a,b,c,d)=(2,4,3,1)$ is one of the solutions.

Case 2: If $d$ is an even integer.
Notice that $b$ is an even integer and $5\mid5^{d-1}\mid b-2d$ , that is; $b-2d\geqslant 6$ .
We may rewrite our main equation as
$5^c-3^b=2^{2c-2d}5^d-3^{2d}2^{b-2d}.$ Now, split into three smaller cases:
(2.1) If $b=2c$ , then $9^c-5^c=2^{2c-2d}(9^d-5^d)$ .
Since $c>d$ , and by Zsigmondy's theorem, there exists an odd prime number $p\mid 9^c-5^c$ , but $p\nmid 9^d-5^d$ . Which results in a contradiction.

(2.2) If $b>2c$ , then
$5^{c-d}-4^{c-d}<(5^{1/2})^{b-2d}-(4^{1/2})^{b-2d}<3^{b-2d}-2^{b-2d}.$ Which immediately yields
$3^{2d}\left(3^{b-2d}-2^{b-2d}\right)>5^d\left(5^{c-d}-4^{c-d}\right).$ Hence, it is impossible to have $b>2c$ .

(2.3) If $b<2c$ , then obviously $\nu_2\left(2^{2c-2d}5^d-3^{2d}2^{b-2d}\right)=b-2d$ . And,
$\nu_2(5^c-3^b)=\nu_2\left((5^c-1)-(3^b-1)\right)\geqslant \min\{2+\nu_2(c),2+\nu_2(b)\}.$ If $\nu_2(b)<\nu_2(c)$ , then $\nu_2(b)=b-2d-2$ . Since $b-2d=5^{d-1}k$ for some positive integer $5\nmid k$ ,
$\nu_2\left(5^{d-1}k+2d\right)=5^{d-1}k-2.$ But, it is false because
$5^{d-1}k-2>\log_2\left(5^{d-1}k+2d\right)\geqslant \nu_2\left(5^{d-1}k+2d\right)$ for every positive integer $d\geqslant 2$ and $k$ .
If $\nu_2(b)>\nu_2(c)$ , then $\nu_2(c)=b-2d-2<\nu_2(b)$ . We may write $\nu_2(b)=b-2d+x-2$ for some positive integer $x$ for some odd positive integer $k$ . Similarly,
$\nu_2\left(5^{d-1}k+2d\right)=5^{d-1}k+2d+x-2.$ It is still false because
$5^{d-1}k+x-2>5^{d-1}k-1>\log_2\left(5^{d-1}k+2d\right)\geqslant \nu_2\left(5^{d-1}k+2d\right)$ for every positive integer $d\geqslant 2$ and $k$ .
Therefore, $\nu_2(b)=\nu_2(c)=\lambda$ . We may write $b=2^\lambda B$ and $c=2^{\lambda} C$ for some odd positive integers $B,C$ . Checking $\nu_2$ again,
$\nu_2\left(5^c-3^b\right)=\nu_2\left(5^{2^{\lambda}C}-3^{2^{\lambda}B}\right)=\nu_2(25^C-9^B)+\lambda-1=b-2d.$ and, $\nu_2(25^C-9^B)=\nu_2(5^C-3^B)+\nu_2(5^C+3^B)$ , but $5^C-3^B\equiv 1-(-1)\equiv 2\pmod{4}$ .
Therefore, $\nu_2(5^c-3^b)=\nu_2(5^C+3^B)+\lambda.$ Notice that
$2^{b-2d-\lambda}\leqslant 5^C+3^B.$ Suppose on the contrary that $c>\frac{4}{5}b$ . Consequently,
$5^d\left(5^{c-d}-4^{c-d}\right)>5^d\left(5^{\frac{4}{5}b-d}-4^{\frac{4}{5}b-d}\right).$ Let $\alpha$ be the ratio $b/d$ . Define a function
$f(\alpha)=5^d\left(5^{\frac{4}{5}\alpha d-d}-4^{\frac{4}{5}\alpha d-d}\right)-9^d\left(3^{\alpha d-2d}-2^{\alpha d-2d}\right)$ Claim. $f(\alpha)>0$ for every positive integer $d$ .
Proof

Since, $f(\alpha)>0$ . It contradicts the main equation, therefore, $c\leqslant \frac{4}{5}b$ .
Recall the inequality $2^{b-2d-\lambda}\leqslant 5^C+3^B$ . Observe that $b<2c\implies B<2C$ :
$2^{b-2d-\lambda}\leqslant 5^C+9^C=9^C\left(\left(\frac{5}{9}\right)^C+1\right)<2\cdot 9^C.$ Taking $\log_2$ both sides gives
$\begin{align*} b-2d-\lambda\leqslant 1+C\log_2(9)&\implies d\geqslant \frac{b-C\log_2(9)-1-\lambda}{2}\\ &\implies d\geqslant \frac{2^\lambda \left(\frac{5}{4}C\right)-C\log_2(9)-1-\lambda}{2}\\ &\implies d> \frac{2^{\lambda-1}C+\lambda-6}{2} \end{align*}$ Which is obviously true because
$b\geqslant \frac{5}{4}c,\quad \frac{5}{4}(2^{\lambda}C)-C\log_2(9)-\lambda> 2^{\lambda-1}C-5+\lambda.$ In other way, we may find the upper bound of $d$ by using:
$5^{d-1}-\lambda\leqslant b-2d-\lambda<1+C\log_2(9).$ Which yields
$d< \log_5(1+C\log_2(9)+\lambda)+1.$ Combining both upper bound and lower bound of $d$ ,
$\frac{2^{\lambda-1}C+\lambda-6}{2}<\log_5(1+C\log_2(9)+\lambda)+1.$ The inequality is indeed false for every odd integer $C\geqslant 13$ (as $\lambda\in\mathbb{N}$ varies).
If $C=1$ , $\lambda\in\{1,2,3\}$ and $d=2$ . We omitted $\lambda=1$ , otherwise $c=2=d$ . But, for $\lambda\in\{2,3\}$ , we can't have $9\mid c-2$ .
If $C=3$ , $\lambda\in\{1,2\}$ . Trying both possibilities also fail to $3^{2d-2}\mid c-d$ .
If $11\geqslant C\geqslant 5$ , $\lambda=1$ , and $d$ can only be equal to $2$ . Checking $3^{2}\mid c-2$ ,
$2C\equiv 2\pmod{9}\implies C\equiv 1\pmod{9}.$ Therefore $C=10$ , but, it is still impossible since $C$ should be an odd integer.

In conclusion, the only solutions satisfying the condition are
$\boxed{(a,b,c,d)\in\{(m,m,n,n),(2,4,3,1),(4,2,1,3)\}}$ for every positive integers $m$ and $n$ .

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