Quick Help
Posted: Sat Mar 06, 2021 10:38 am
Let $a,b,c$ three prime numbers satisfying $5\leq a<b<c$ , $2a^2 -c^2 \geq 49$ and $2b^2 -c^2 \leq 193$ what is the biggest $c$?
$\{a,b,c\}=\{17,19,23\}$where c is maximum.
Now we got 2 inequalities,
$49 \leq 2a^2-c^2 \cdots (1)$
and
$2b^2 \leq 193 +c^2 \cdots (2)$
By adding (1) and (2) we get,
$2b^2 + 49 \leq 2a^2+193$
Or,$2b^2 \leq 2a^2 + 144$
Or,$b^2 \leq a^2 + 72$
Or,$b^2 -a^2 \leq 72 $
Now, this narrows down or set of possible outcomes a lot.
$(b+a)(b-a) \leq 72$
So the maximum possible value of $b+a$ is 36 because if it is more, the product must be more than 72 as the minimum value of $b-a$ is 2[No prime other than 2,3 are consequitive]
Now our suspect list is with primes less than 23. Only these prime can be a or b.
Now we can use trial and error to find the max value of c is 23.$\square$
Now we got 2 inequalities,
$49 \leq 2a^2-c^2 \cdots (1)$
and
$2b^2 \leq 193 +c^2 \cdots (2)$
By adding (1) and (2) we get,
$2b^2 + 49 \leq 2a^2+193$
Or,$2b^2 \leq 2a^2 + 144$
Or,$b^2 \leq a^2 + 72$
Or,$b^2 -a^2 \leq 72 $
Now, this narrows down or set of possible outcomes a lot.
$(b+a)(b-a) \leq 72$
So the maximum possible value of $b+a$ is 36 because if it is more, the product must be more than 72 as the minimum value of $b-a$ is 2[No prime other than 2,3 are consequitive]
Now our suspect list is with primes less than 23. Only these prime can be a or b.
Now we can use trial and error to find the max value of c is 23.$\square$