Quick Help

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Asif Hossain
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Joined:Sat Jan 02, 2021 9:28 pm
Quick Help

Unread post by Asif Hossain » 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 $c$?
Hmm..Hammer...Treat everything as nail

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Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: Quick Help

Unread post by Mehrab4226 » Sat Mar 06, 2021 11:22 pm

$\{a,b,c\}=\{17,19,23\}$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.$\square$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Quick Help

Unread post by Asif Hossain » Sun Mar 07, 2021 9:15 am

Mehrab4226 wrote:
Sat Mar 06, 2021 11:22 pm
$\{a,b,c\}=\{17,19,23\}$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.$\square$
Is there any systematic approach rather than any trial and error??
Hmm..Hammer...Treat everything as nail

User avatar
Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: Quick Help

Unread post by Mehrab4226 » Sun Mar 07, 2021 7:45 pm

Asif Hossain wrote:
Sun Mar 07, 2021 9:15 am
Mehrab4226 wrote:
Sat Mar 06, 2021 11:22 pm
$\{a,b,c\}=\{17,19,23\}$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.$\square$
Is there any systematic approach rather than any trial and error??
Actually what I did is find all the solutions. A better approach may be possible. But there aren't that many solutions so....
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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