A number theory problem by A Former Brilliant Member

We only need clues (1) and (4):

Say the numbers are $a,a+d,a+2d,a+3d,a+4d$ .

Clue (1): $2a+d=a+4d$ , so we have $a=3d$ and we can rewrite the list as $3d,4d,5d,6d,7d$ .

Clues (2) and (3) follow directly from this; they're the relations $3^2+4^2=5^2$ and $3^3+4^3+5^3=6^3$ respectively.

Clue (4): $25d=25d^2$ ; since the integers are positive, $d=1$ and the sum is $\boxed{25}$ .

Logical (and lucky) approach:

Let's say we have an arithmetic sequence: $a_{1}, a_{2},a_{3},a_{4},a_{5}$ .

The second clue implies that $a_{1}^{2}+a_{2}^{2}=a_{3}^{2}$ , means that the first three numbers form a Pythagoras triple. When I search for the satisfied one, the trio $(3,4,5)$ do form a arithmetic sequence (1). To check again, let's see the first clue, states that $a_{1}+a_{2}=a_{5}$ . Series (1) has common difference of $1$ , so $a_{5}=7$ , which satisfy the condition.

So the sequence is $\boxed{3,4,5,6,7}$ with $Sum=\boxed{25}$ .

$a_1+4d=a_2+3d=a_3+2d=a_4+d=a_5$

• $a_1+a_2=a_5\implies a_1=3d$
• $a_1 ^2+a_2 ^2=a_3 ^2\\ 9d^2+16d^2=25d^2$ - This is true for any $d$
• $a_1 ^3+a_2 ^3+a_3 ^3=a_4 ^3\\ 27d^3+64d^3+125d^3=216d^3$ - This is true for any $d$
• $(3+4+5+6+7)d=25d^2\implies d=1$

The solution is $3+4+5+6+7=\boxed{25}$

Suppose the terms are $a-2d, a-d, a, a+d, a+2d$ . We can get the answer just by the 4th clue.

Now, as per the 4th clue,

$a-2d + a-d + a + a+d+a+2d={ a }^{ 2 }$

$5a={ a }^{ 2 }$

Now, dividing both sides by $a$ as it is not 0 as all are positive integers.

$5=a$

Thus, the sum of all the terms would be

$= a-2d + a-d + a + a+d+a+2d$

$=5a$

$=5 \times 5 \quad .... \quad (\text{As we found} \quad a=5)$

Thus, the sum is $5 \times 5=25$ .