(1 + 2 n) (1 + 2 m) = 1 + 2 ( m + n + 2 m n) = 99 = 1 * 99 = 3 * 33 = 9.* 11 Case I. ( 1 + 2 m. = 1, 1 + 2 n. = 99 Or 1 + 2 n. = 1, 1 + 2 m. = 99 ) Hereby m = 0, n. = 49 or m = 49, n = 0 Therefore, m + n = 49 Case II ( 1 + 2 m = 3, 1 + 2 n. = 33 Or 1 + 2 n. = 33, 1 + 2 m. = 3 ) Hereby m = 1, n. = 16 or m = 16, n = 1 Therefore, m + n = 17 Case III ( 1 + 2 m = 9, 1 + 2 n. = 11 Or 1 + 2 n. = 9, 1 + 2 m. = 11 ) Hereby m = 4, n. = 5 or m = 5, n = 4 Therefore, m + n = 9
As there are 2 unknowns m and n m+2mn+n=49 is a diophantine equation --> m and n are integers 49=±1×(±49) or 49=(±7)² 49=(±7)² is impossible as LHS is not a square. If m=1 then 49=1+2n+n --> n=16 If m=49 then 49=49+2×49n+n 0=99n --> n>0 m=-1 or m=-49 does not aplly. As the equation is cyclical then (m,n)={(1,16),(0,49),(16,1),(49,0)}
This is, again, a very complicated approach: Simply setting m to 1, then trivially calculating n (it's the most simple linear expression that exists) gets you directly to the first pair, then going up from there as long as m is lower than n yields another 3 trivial cases. Since the expression is symmetric, the solutions are as well. Then the thing is solved. I don't see how "90% failed to solve" could possibly fit into reality. Could it be that this wording is just clickbait?
Have you worked out your method in its entirety? Because finding out what values of m you need to plug in sounds pretty painful. (On top of that, the method he presented also works for all Diophantine equations of the form am+bmn+cn=d)
@@grrgrrgrr0202 There was nothing "painful" about the trivial approach. It had just 4 pairs of (m,n) to examine - MUCH LESS than what the author of the video presented in his approach. And AFTER having his complicated reforming of expressions, he ended up probing pairs of values anyways - he didn't reduce complexity at ANY stage, just to the contrary.
@@WhiteGandalfsYour method may be preferable in this case, but it would fall pretty flat if the right hand number was bigger or if m and n didn't have any coefficients. It is also worth noting that TH-cam videos like these go for extremely slow and overly detailed explanations (which I despise); the method listed here takes arguably less than 30 seconds to resolve.
(1 + 2 n) (1 + 2 m)
= 1 + 2 ( m + n + 2 m n)
= 99 = 1 * 99 = 3 * 33
= 9.* 11
Case I. ( 1 + 2 m. = 1, 1 + 2 n. = 99
Or 1 + 2 n. = 1, 1 + 2 m. = 99 )
Hereby m = 0, n. = 49
or m = 49, n = 0
Therefore, m + n = 49
Case II ( 1 + 2 m = 3, 1 + 2 n. = 33
Or 1 + 2 n. = 33, 1 + 2 m. = 3 )
Hereby m = 1, n. = 16
or m = 16, n = 1
Therefore, m + n = 17
Case III ( 1 + 2 m = 9, 1 + 2 n. = 11
Or 1 + 2 n. = 9, 1 + 2 m. = 11 )
Hereby m = 4, n. = 5 or m = 5, n = 4
Therefore, m + n = 9
i did a hard push; n=1 then m=16; 2&3 then m not integer; n=4 then m=5 .
As there are 2 unknowns m and n m+2mn+n=49 is a diophantine equation --> m and n are integers
49=±1×(±49) or 49=(±7)²
49=(±7)² is impossible as LHS is not a square.
If m=1 then 49=1+2n+n --> n=16
If m=49 then 49=49+2×49n+n
0=99n --> n>0
m=-1 or m=-49 does not aplly.
As the equation is cyclical then
(m,n)={(1,16),(0,49),(16,1),(49,0)}
Nice👍
Great explaination !!
Thank you ⚘️
It's easy to solve mentaly,
M = 1 or 16
N = 16 or 1
Now, I'll try to get a arithmetic solution ...
Thank you sir for such good questions and explanations
Thank you
👏👏👏
Thanks for watching !!
m+2mm+n=√49 (m+7n-7)
This is, again, a very complicated approach: Simply setting m to 1, then trivially calculating n (it's the most simple linear expression that exists) gets you directly to the first pair, then going up from there as long as m is lower than n yields another 3 trivial cases. Since the expression is symmetric, the solutions are as well. Then the thing is solved. I don't see how "90% failed to solve" could possibly fit into reality. Could it be that this wording is just clickbait?
Have you worked out your method in its entirety? Because finding out what values of m you need to plug in sounds pretty painful. (On top of that, the method he presented also works for all Diophantine equations of the form am+bmn+cn=d)
@@grrgrrgrr0202 There was nothing "painful" about the trivial approach. It had just 4 pairs of (m,n) to examine - MUCH LESS than what the author of the video presented in his approach. And AFTER having his complicated reforming of expressions, he ended up probing pairs of values anyways - he didn't reduce complexity at ANY stage, just to the contrary.
Ok, so how would you figure out what values of m or n you'd need to examine?
@@grrgrrgrr0202 The expression is symmetric to m and n, thus you need only values m
@@WhiteGandalfsYour method may be preferable in this case, but it would fall pretty flat if the right hand number was bigger or if m and n didn't have any coefficients. It is also worth noting that TH-cam videos like these go for extremely slow and overly detailed explanations (which I despise); the method listed here takes arguably less than 30 seconds to resolve.
asnwer=7 isit