Seven Haven

As most would have realized, by testing small values, $a=10$ seems to be the only solution. However, it is harder to show that this is indeed the only possible solution.

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First of all, if $a$ is odd, then both $a^2$ and $2^a-a^3$ are odd, so $a^2(2^a-a^3)+1$ is even, thus it is not a power of $7.$ So $a=2c$ for some integer $c.$

Suppose $a^2(2^a-a^3)+1=7^b$ . Replacing $a$ by $2c,$ we get $4c^2(4^c-8c^3)=7^b-1$ Modulo $3,$ we get $c^2(1+c)\equiv 0\pmod 3,$ so either $c\equiv 0$ or $c\equiv 2$ modulo $3.$

Now we consider this equation modulo $7$ . We know that $4^3 \equiv 1 \pmod{7}$ . Note that if $c\equiv 0 \pmod 3,$ then $4^c\equiv 1 \pmod 7;$ if $c\equiv 2 \pmod 3,$ then $4^c\equiv 2 \pmod 7.$ Looking at all different residues of $c$ modulo $7$ we get that the only possibility is that $c\equiv 2\pmod 3$ and $c\equiv 5 \pmod 7.$ This means $c\equiv 5 \pmod {21},$ so $c=5,26,47,...$ It is easy to check that $c=5$ works, and it corresponds to $a=10.$ We will prove that this is the only possibility.

Now considering the same equation modulo $8,$ we easily conclude that $b$ is even, say $b=2d.$ Then note that $4c^2\cdot 4^c$ and $7^b=7^{2d}$ are both perfect squares. Because for $c\geq 26$ , $4c^2\cdot (4^c-8c^3)+1$ is less than $4c^2\cdot 4^c,$ it must be less than or equal to $(2c\cdot 2^c-1)^2.$ This implies $32c^5\geq 2\cdot 2c\cdot 2^c$ . This simplifies to $2^c\leq 8c^4$ It is easy to see that this inequality is false for $c\geq 26,$ thus $c=5$ is the only solution.

So the answer is 10.

First observe that $a$ must be even. This means

$a^2\left(2^a-a^3\right)+1\equiv1\pmod{3}\implies a\equiv0,1\pmod{3}\implies a\equiv0,4\pmod{6}$

For $a\equiv0\pmod{6}$ , we see that this is impossible after plugging in $a=6k$ and testing all $0\le k\le6$ modulo $7$ . For $a\equiv4\pmod{6}$ , we see that $a=6k+4$ is only a residue for $k=7k+1$ modulo $7$ . Hence, $a\equiv10\pmod{42}$ . But we see that if $k\ge1$ , then $a=42k+10$ is a nonresidue, meaning that $k=0$ is the only solution: $\boxed{10}$ .

a^2(2^a - a^3) + 1 = 7^n.

a^2(2^a - a^3) = 7^n - 1.

7^n - 1 is always even.

=> a^2(2^a - a^3) is also always even.

2^a is even so if a were odd, odd x (even - odd) =odd x odd = odd.

and thats bad.

so a is even.

as 7^n is exponential is is never negative.

and 7^n - 1 is positive ( provided n > 0 ).

so we want a^2(2^a - a^3) to be positive.

for that 2^a - a^3 > 0.

2^a > a^3.

and this happens when a =< 10.

so now we have all tools ready, we are going to brute now.

in a^2(2^a - a^3) + 1 = 7^n.

a=10 then a^2(2^a - a^3) + 1 = 10^2(2^10 - 10^3) + 1 = 100(1024 - 1000) + 1 = 2401 = 7^4.

and for a > 10 we get no value as seven exponentials.

we have only a=10 as solution.

precisely

∑a = 10.

a=10 is the only value which equals a power of 7; a=10 gives 2401 which is 7^4

for the above equation to be true 2^1 must be greater than a^3 . so if u check for values of a for which the above eqn is true ull find that a>=10 satisfies the eqn . so substituting a=10 we get 2401=7^3 The PROBLEM CAN ALSO BE SOLVED IN ANOTHER METHOD power of 7's -->units digit-->can be ----->1,7,3,9 and so we can conclude sayin that a^2(2^a-a^3) is even becoz wen u subtract the 1 in the ques with the units digit we infer that its even so a^2 has to be even becoz if a^2 is odd then a^3 is also odd and we will get a even quantity which can never be a power of 7. Therefore 2^a-a^3 shud be even Checkin for values only possibility a =10 hence PROVED ;)