A Fresh Power Diophantine Equation

It is clear that $a>b \ge 1$ . Looking mod $b$ gives $b| a^{n-1} (a-1)$ and since $\gcd(a, b)=1$ we get $b|a-1$ and $a=bc+1$ for natural number $c$ . Looking mod $a$ gives $a|b^3+1$ . Thus $a|b^3+1-a=b^3+1-(bc+1)=b(b^2-c)$ and there is a non-negative integer $d$ such that $b^2-c=da=d(bc+1)$ or $b^2-cdb-(c+d)=0$ Which is a quadratic in $b$ . Its discriminant must be a perfect square, but note that for $c>1$ and $d>1$ $(cd)^2 \le \Delta_b=c^2d^2+4(c+d) \le (cd+2)^2$ Thus we must have $c^2d^2+4(c+d)=(cd+1)^2$ , which leads to $c=\frac{4d-1}{2d-4}$ Which is not an integer. This is a contradiction and so $c=1$ or $d \le 1$ . Therefore we have three cases to check: $c=1$ , $d=0$ and $d=1$ .

For $c=1$ we get $a=b+1$ and equation becomes $(b+1)^n=b^{n+2}+(2b+1)^{n-1} \qquad (1)$ From this equation observe that $(b+1)^n > b^{n+2}$ and $(b+1)^n > (2b+1)^{n-1}$ . Multiply these two inequalities to get $(b^2+2b+1)^n > (2b^2+b)^{n-1} b^3$ But for $b \ge 3$ we have $(b^2+b+1)^n=(b^2+b+1)^{n-1} (b^2+b+1)<(2b^2+b)^{n-1} b^3$ which is a contradiction and $b \le 2$ . For $b=1$ from equation $(1)$ we get $2^n=1+3^{n-1}$ and one can prove by induction that $1+3^{n-1}>2^n$ for $n>2$ . $n=2$ satisfies the equation and leads to $a=2$ . We get a solution from here: $(a, b, n)=(2, 1, 2)$ . For $b=2$ from equation $(1)$ we get $3^n=2^{n+2}+5^{n-1}$ and one can prove by induction that $3^n<2^{n+2}+5^{n-1}$ for $n>3$ and for $n \le 3$ equality does not hold.

For $d=0$ we have $a=b^3+1$ and original equation becomes $(b^3+1)^n-b^{n+2}=(b^3+b+1)^{n-1} \qquad (2)$ Suppose that $p$ is prime divisor of $b^3+b+1$ . Looking mod $p$ gives $‎(-b)^n ‎‎\stackrel{‎p‎}{\equiv} b^{n+2}$ or $‎(-1)^n ‎‎\stackrel{‎p‎}{\equiv} b^2$ . For odd $n$ it follows that $‎b^2+1 ‎‎\stackrel{‎p‎}{\equiv} 0$ , which is impossible considering that $\gcd(b^2+1, b^3+b+1)=1$ . Thus $n$ is even and $p| b^2-1$ and $p|b^3+b+1-b(b^2-1)=2b+1 \\ p|b(2b+1)-2(b^2-1)=b+2 \\ p|2(b+2)-(2b+1)=3$ So $p=3$ is the only prime divisor of $b^3+b+1$ and $b^3+b+1=3^k$ for a natural number $k$ . Suppose that $k>1$ . RHS of $b^3+b+1=3^k$ is divisible by $9$ . Let us look when $b^3+b+1$ is divisible by $9$ . \begin{array}{|*{10}{c|}} \hline b \pmod 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline b^3 \pmod 9 & 0 & 1 & 8 & 0 & 1 & 8 & 0 & 1 & 8 \\ \hline b^3+b+1 \pmod 9 & 1 & 3 & 2 & 4 & 6 & 5 & 7 & 0 & 8 \\ \hline \end{array} Thus we must have $b \equiv 7 \pmod{9}$ . Now from equation $(2)$ looking mod $9$ gives $2^n-7^{n+2} ‎‎\stackrel{‎9‎}{\equiv} 0$ , which is impossible. Thus $k=1$ and $b=1$ and $2^n-1=3^{n-1}$ , which gives the same solution of case $c=1$ .

Final case is $d=1$ , which gives $a=b^2-b+1$ and equation becomes $(b^2-b+1)^n-b^{n+2}=(b^2+1)^{n-1} \qquad (3)$ Suppose that $p$ is prime divisor of $b-1$ . Looking mod $p$ gives $‎0 ‎‎\stackrel{‎p‎}{\equiv} 2^{n-1}$ and tells us that $2$ is the only prime divisor of $b-1$ and $b=2^k+1$ for a natural number $k$ . Once again looking mod $b^2+1$ in equation $(3)$ gives $(-b)^n \equiv b^{n+2} \pmod{b^2+1}$ . For even $n$ it follows that $‎b^2-1 \equiv 0 \pmod{b^2+1}$ , which is impossible and thus $n$ is odd. Finally look mod $2^k$ in equation $(3)$ ; It follows that $0 \equiv 2^{n-1} \pmod{2^k}$ , thus $n-1 \ge k$ . Now rearrange equation $3$ in the following form $(b^2-b+1)^n-b^n=b^n(b^2-1)+(b^2+1)^{n-1} \qquad (4)$ Suppose $k>1$ . Let us find power of $2$ in the LHS of this equation using LTE lemma. Note that since $n$ is odd $‎v_2 \left(‎‎(b^2‎-b‎+1)^n-b^n \right)‎=‎v_2(b^2-b+1-b)+v_2(n)+‎v_2(b^2+1)-1‎‎=2k \qquad (5)$ For finding power of $2$ in the RHS of equation $(5)$ write $\displaystyle \begin{array}{c}‎ ‎v_2 \left(b^n(b^2-1)+(‎b^2‎+1)^{n-1} \right) &=‎\min \left\{ ‎‎v_2\left(b^n(b^2-1) \right), ‎‎v_2\left((‎b^2‎+1)^{n-1} \right)\right\} \\ &=\min \left\{‎‎v_2 \big( 2^{k+1} (2^{k-1}+1)‎(2^k+1)^n \big), ‎‎v_2\left(2^{n-1}(2^{2k-1}+2^k+1)^{n-1} \right)\right\} \\ &=\min \left\{k+1, n-1\right\} \\ &=\frac{1}{2}(k+n-|k-n+2|) \end{array}$ So from relation $(5)$ $2k=\frac{1}{2}(k+n-|k-n+2|) \qquad (6)$ But we proved $n-1 \ge k$ ; If $k-n=-1$ relation $(6)$ gives $2k=k$ , which is impossible and for $k-n \le -2$ relation $(6)$ turns to $2k=\frac{1}{2}(k+n-|k-n+2|)=\frac{1}{2}(k+n+k-n+2)=‎k+1$ Which gives $k=1$ , but this is a contradiction, because we supposed $k>1$ . So $k=1$ and $b=3$ . Putting these values in equation $(3)$ gives $7^n=3^{n+2}+10^{n-1}$ . One can prove using induction that $7^n<3^{n+2}+10^{n-1}$ for $n>6$ . Trying values $n \le 6$ gives $n=3$ as an answer. It leads to second solution: $(a, b, n)=(7, 3, 3)$ .

The solution is very lengthy. Here's the solution to the '97 Russian Olympiad problem: $p^{3} - q^{5} = (p + q)^{2}$

First suppose neither $p$ nor $q$ equals 3. If they are congruent modulo 3, the left side is divisible by 3 while the right is not; if they are not congruent modulo 3, the right side is divisible by 3 while the left side is not, so there are no such solutions. If $p = 3$ , we have $q^{5} < 27$ , which is impossible. Thus $q= 3$ and $p^{3} - 243 = (p+ 3)^{2}$ , whose only integer root is $p = 7$ .

For the generalised problem $a^{n} - b^{n+2} = (a + b)^{n-1}$ , we can observe another trivial solution for $n = 2$ which is $(2,1,2)$ . If we can prove that there exists no solution for $n \geq 4$ , then we have our answer here as $(7+3+3)+(2+1+2) = 18$