
NT5. Find all the positive integers $x, y, z, t$ such that $2^{x} \cdot 3^{y}+5^{z}=7^{t}$.


Solution. Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$. .

Next we prove that $t$ is even. Obviously, $t \geq 2$. Let us suppose that $t$ is odd, $t=2 d+1$, $d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$.

If $x \geq 2$, reducing modulo 4 , we get $1 \equiv 3$, contradiction.

For $x=1$, we have $2 \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$; and, reducing modulo 24 , we obtain $2 \cdot 3^{y}+1 \equiv$ $7 \Rightarrow 24 \mid 2\left(3^{y}-3\right)$, i.e. $4 \mid 3^{y-1}-1$, which means that $y-1$ is even. Then, $y=2 b+1$, $b \in \mathbb{N}$.

We obtain $6 \cdot 9^{b}+25^{c}=7 \cdot 49^{d}$, and, reducing modulo 5 , we get $(-1)^{b} \equiv 2 \cdot(-1)^{d}$, which is false, for all $b, d \in \mathbb{N}$. Hence $t$ is even, $t=2 d, d \in \mathbb{N}$. .

The equation can be written as $2^{x} \cdot 3^{y}+25^{c}=49^{d} \Leftrightarrow 2^{x} \cdot 3^{y}=\left(7^{d}-5^{c}\right)\left(7^{d}+5^{c}\right) . \quad$. As $\operatorname{gcd}\left(7^{d}-5^{c}, 7^{d}+5^{c}\right)=2$ and $7^{c}+5^{c}>2$, there exist exactly three possibilities:(1)
$\left\{\begin{array}{l}7^{d}-5^{c}=2^{x-1} \\ 7^{d}+5^{c}=2 \cdot 3^{y}\end{array}\right.$
(2) $\left\{\begin{array}{l}7^{d}-5^{c}=2 \cdot 3^{y} \\ 7^{d}+5^{c}=2^{x-1}\end{array}\right.$
(3) $\left\{\begin{array}{l}7^{d}-5^{c}=2 \\ 7^{d}+5^{c}=2^{x-1} \cdot 3^{y}\end{array}\right.$

Case (1). We have $7^{d}=2^{x-2}+3^{y}$ and, reducing modulo 3 , we get $2^{x-2} \equiv 1(\bmod 3)$, hence $x-2$ is even, i.e. $x=2 a+2$, where $a \in \mathbb{N}$, since $a=0$ would mean $3^{y}+1=7^{d}$, which is impossible (even $=$ odd).

We obtain $7^{d}-5^{c}=2 \cdot 4^{a} \stackrel{\text { mod } 4}{\Longrightarrow} 7^{d} \equiv 1(\bmod 4) \Rightarrow d=2 e, e \in \mathbb{N}$. Then $49^{e}-5^{c}=$ $2 \cdot 4^{k} \stackrel{\bmod 8}{\Longrightarrow} 5^{c} \equiv 1(\bmod 8) \Rightarrow c=2 f, f \in \mathbb{N}$. We obtain $49^{e}-25^{f}=2: 4^{a} \xlongequal{\bmod 3} 0 \equiv 2(\bmod 3)$, false. In conclusion, in this case there are no solutions to the equation.

Case (2). From $2^{x-1}=7^{d}+5^{c} \geq 12$, we obtain $x \geq 5$. Then $7^{d}+5^{c} \equiv 0(\bmod .4)$, i.e. $3^{d}+1 \equiv 0(\bmod 4)$, hence $d$ is odd. As $7^{d}=5^{c}+2 \cdot 3^{y} \geq 11$, we get $d \geq 2$, hence $d=2 e+1$, $e \in \mathbb{N}$.

As in the previous case, from $7^{d}=2^{x-2}+3^{y}$, reducing modulo 3, we obtain $x=2 a+2$, with $a \geq 2$ (because $x \geq 5$ ). We get $7^{d}=4^{a}+3^{y}$, i.e. $7 \cdot 49^{e}=4^{a}+3^{y}$, hence, reducing modulo 8 , we obtain $7 \equiv 3^{y}$, which is false, because $3^{y}$ is congruent mod8 either to 1 (if $y$ is even) or to. 3 (if $y$ is odd). In conclusion, in this case there are no solutions to the equation. $k \in \mathbb{N}$.

Case (3). From $7^{d}=5^{c}+2$, it follows that the last digit of $7^{d}$ is 7 , hence $d=4 k+1$, false.

If $c \geq 2$, from $7^{4 k+1}=5^{c}+2$, reducing modulo 25 , we obtain $7 \equiv 2(\bmod 25)$, which is

For $c=1$ we get $d=1$, and the solution $x=3, y=1, z=t=2$.

3 NT6. If $a, b, c, d$ are integers and $A=2(a-2 b+c)^{4}+2(b-2 c+a)^{4}+2(c-2 a+b)^{4}$, $B=d(d+1)(d+2)(d+3)+1$, prove that $(\sqrt{A}+1)^{2}+B$ cannot be a perfect square.
