Does anybody have solution?

We have $(a^3+b)(b^3+a)=2^c$ and since $3 \nmid 2^c$ then it is easy to prove that $c$ is even. Case 1. $a=b$ then $a^2(a^2+1)^2=2^c$ implies $a=b=1$. Case 2. $a \ne b$. WLOG $a>b \ge 1$. If $a=1$ then $(b+1)^2(b^2-b+1)=2^c=2^{2c_1}$. Since $b \ge 1$ so $b^2-b+1$ must be a perfect square. It follows that $b=1$. Similarly, if $b=1$ then $a=1$. If $a \ge 2$ then $b \ge 2$. Therefore $2^c \ge 100$ or $c \ge 7$. It is easy to prove that $\gcd (a,b)=1$ and $2 \nmid a, 2 \nmid b$. The equation is equivalent to \[ab(ab-1)^2+(a^2+b^2)^2=2^c.\] Since $a,b \equiv 1 \pmod{2}$ then $a^2+b^2 \equiv 2 \pmod{4}$. Hence $(a^2+b^2)^2 \equiv 4 \pmod{8}$. But $c \ge 7$ so $16|2^c$. Therefore $ab(ab-1)^2 \equiv 4 \pmod{8}$ or $(ab-1)^2 \equiv 4 \pmod{8}$. Thus, $ab \equiv 3 \pmod{4}$. It follows that $a-b \equiv 2 \pmod{4}$. Let $a^3+b=2^m,b^3+a=2^n \; (c > m> n \ge 1, m+n=c)$ then \[2^n \left( 2^{m-n}-1 \right)=(a-b)(a^2+b^2+ab-1).\] Since $2 \parallel a-b$ then $2^{n-1} \parallel a^2+ab+b^2-1$ or $2^n \parallel 2(a^2+ab+b^2-1)$. It follows that $a+b^3|2(a^2+ab+b^2-1)$. We get $a+b^3|2b(a^2+ab+b^2-1)$ or $a+b^3|2[(ab-1)(a+b)+a+b^3]$. Hence $a+b^3|2(ab-1)(a+b)$. But we also have $ab \equiv 2 \pmod{4}$ so $a+b^3|4(a+b)$. Let $\frac{4(a+b)}{a+b^3}=k$ with $k \in \mathbb{N},k \ge 1, 2 \nmid k$. Therefore $a= \frac{b(b^2k-4)}{4-k}$. This implies $k \le 4$. We get $k=1$ or $k=3$. If $k=1$ then $a= \frac{b^3-4b}{3}$. Therefore $2^{n-2}= \frac{b(b-1)(b+1)}{3}$. From here we get $b=3$. Therefore $a=5$. If $k=3$ then $a= b(3b^2-4)$. We obtain $2^n=a+b^3=4(b-1)b(b+1)$. We find $b=1$, a contradiction since $b \ge 2$. Thus, the answer is $(a,b)=(1;1),(3;5),(5;3)$. $\blacksquare$

Here is an elegant solution which is result of several days working. I hope its true. Its easy to see that $a,b$ are odd and relatively prime. If $a=b$ we have $a=b=1$ WLOG assume that $a>b$. We have $2^x=a^3+b>b^3+a=2^y$ which leads $x>y$ so $b^3+a|a^3+b$ => $b^3+a|b^9-b$ => $b^3+a|b^8-1=(b^4+1)(b^2+1)(b^2-1)$ Since $2||b^4+1,b^2+1$ we have $2^{y-2}|b^2-1$ which means $4(b^2-1)>=2^y$ but we also have $2^y>b^3$ . So we have $4(b^2-1)>b^3$ or equivalently $b=2,3$ . Rest is easy...

shinichiman wrote: Therefore $ab(ab-1)^2 \equiv 4 \pmod{8}$ or $(ab-1)^2 \equiv 4 \pmod{8}$. Thus, $ab \equiv 3 \pmod{4}$. It follows that $a-b \equiv 2 \pmod{4}$. Or $ab=7mod8$

Αρχιμήδης 6 wrote: shinichiman wrote: Therefore $ab(ab-1)^2 \equiv 4 \pmod{8}$ or $(ab-1)^2 \equiv 4 \pmod{8}$. Thus, $ab \equiv 3 \pmod{4}$. It follows that $a-b \equiv 2 \pmod{4}$. Or $ab=7mod8$ What do you mean ?

shinichiman wrote: We have $(a^3+b)(b^3+a)=2^c$ and since $3 \nmid 2^c$ then it is easy to prove that $c$ is even. Case 1. $a=b$ then $a^2(a^2+1)^2=2^c$ implies $a=b=1$. Case 2. $a \ne b$. WLOG $a>b \ge 1$. If $a=1$ then $(b+1)^2(b^2-b+1)=2^c=2^{2c_1}$. Since $b \ge 1$ so $b^2-b+1$ must be a perfect square. It follows that $b=1$. Similarly, if $b=1$ then $a=1$. If $a \ge 2$ then $b \ge 2$. Therefore $2^c \ge 100$ or $c \ge 7$. It is easy to prove that $\gcd (a,b)=1$ and $2 \nmid a, 2 \nmid b$. The equation is equivalent to \[ab(ab-1)^2+(a^2+b^2)^2=2^c.\]Since $a,b \equiv 1 \pmod{2}$ then $a^2+b^2 \equiv 2 \pmod{4}$. Hence $(a^2+b^2)^2 \equiv 4 \pmod{8}$. But $c \ge 7$ so $16|2^c$. Therefore $ab(ab-1)^2 \equiv 4 \pmod{8}$ or $(ab-1)^2 \equiv 4 \pmod{8}$. Thus, $ab \equiv 3 \pmod{4}$. It follows that $a-b \equiv 2 \pmod{4}$. $\blacksquare$ How u are doing the part .. $ab \equiv 3\pmod{4}$

Mathematicalx wrote: Here is an elegant solution which is result of several days working. I hope its true. Its easy to see that $a,b$ are odd and relatively prime. If $a=b$ we have $a=b=1$ WLOG assume that $a>b$. We have $2^x=a^3+b>b^3+a=2^y$ which leads $x>y$ so $b^3+a|a^3+b$ => $b^3+a|b^9-b$ => $b^3+a|b^8-1=(b^4+1)(b^2+1)(b^2-1)$ Since $2||b^4+1,b^2+1$ we have $2^{y-2}|b^2-1$ which means $4(b^2-1)>=2^y$ but we also have $2^y>b^3$ . So we have $4(b^2-1)>b^3$ or equivalently $b=2,3$ . Rest is easy... How did you get $b^3+a|b^9-b$ from $b^3+a|a^3+b$?

Milana wrote: Mathematicalx wrote: Here is an elegant solution which is result of several days working. I hope its true. Its easy to see that $a,b$ are odd and relatively prime. If $a=b$ we have $a=b=1$ WLOG assume that $a>b$. We have $2^x=a^3+b>b^3+a=2^y$ which leads $x>y$ so $b^3+a|a^3+b$ => $b^3+a|b^9-b$ => $b^3+a|b^8-1=(b^4+1)(b^2+1)(b^2-1)$ Since $2||b^4+1,b^2+1$ we have $2^{y-2}|b^2-1$ which means $4(b^2-1)>=2^y$ but we also have $2^y>b^3$ . So we have $4(b^2-1)>b^3$ or equivalently $b=2,3$ . Rest is easy... How did you get $b^3+a|b^9-b$ from $b^3+a|a^3+b$? Since b³+a divides b⁹+a³ and a³+b it divides also b⁹-b

\[(a^2+b^2)^2+ab(ab-1)^2=2^c\]$2^m=a^3+b\equiv a+b^3=2^n(mod \ 3)\implies 2|m+n=c$ We also get that $a,b$ are odd since assuming $v_2(a)\geq v_2(b)+1$ gives contradiction. $i)min\{a,b\}>5$. Let $c=2d$. \[ab(ab-1)^2=(2^d-(a^2+b^2))(2^d+(a^2+b^2))\]Since $a,b$ are odd, we have $2^d\pm (a^2+b^2)\equiv 2(mod \ 4)$ Thus $v_2(RHS)=v_2(LHS)=2\implies v_2(ab-1)=1$ Also $(2^d+a^2+b^2,2^d-a^2-b^2)|2^d\implies \frac{(ab-1)^2}{4}$ divides $2^d+(a^2+b^2)$ or $2^d-(a^2+b^2)$. If $(ab-1)^2|4(2^d-a^2-b^2)$, then $2^d+a^2+b^2|4ab$ \[2^d\leq 4ab-a^2-b^2\leq 2ab\implies ab\geq 2^d\]\[0>2^{d+2}-2^{d+3}\geq 2^{d+2}-8ab\geq 2^{d+2}-4a^2-4b^2\geq (ab-1)^2\]Which results in a contradiction. Hence $(ab-1)^2|4(2^d+a^2+b^2)$ $(ab-1)^2|2^{d+2}+4a^2+4b^2$ $2^d-a^2-b^2|4ab$ \[(ab-1)^2\leq 2^{d+2}+4a^2+4b^2\leq 4a^2+16ab+4b^2+4a^2+4b^2=8(a+b)^2\]Hence $(3a+3b)^2>8(a+b)^2\geq (ab-1)^2\implies 3a+3b>ab-1\implies 10>(a-3)(b-3)\geq 9\implies a=b=6$ which doesn't satisfy. $ii)min\{a,b\}=1$ WLOG $b=1$. We have $(a+1)^2(a^2-a+1)=(a^3+1)(a+1)=2^c$ and $a^2-a+1$ is odd so $a=1\implies \boxed{(1,1)}$ $iii)min\{a,b\}=3$ WLOG $b=3$. We have $(a^3+3)(a+27)=2^c$ $a+27<a^3+3$ since $a>3$. Thus $a+27|a^3+3\iff 2^m=a+27|6560\implies a+27|32\implies a=3\implies \boxed{(5,3),(3,5)}$ $iv)min\{a,b\}=5$ $(a^3+5)(a+125)=2^c$ and $a+125<a^3+5$ thus $a+125|a^3+5\implies 2^m=a+125|5^8-1$ $\implies 7\leq m\leq v_2(5^8-1)=5$ which gives no solution as desired.$\blacksquare$