A Diophantine Equation | m=n/(n+5)

⭐ Join this channel to get access to perks:
#shorts #shortsvideo #shortsyoutube
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

[pelledanasten]

n/(n+5)<1 for all positive n
->m<1. But there is no positive integer less than one, so m can not be a positive integer.

[josukehigashikata]

by observation, since n>0 then :-
n+5 > n
therefore n/(n+5) < 1
therefore m < 1
And that means there are no possible positive integers that satisfy the equation.

[dyneek]

m = n/(n+5); m*n + 5m = n; m*n = n - 5m; m = 1-(5m/n); since m is supposed to be an integer. This means 5m/n is also an integer. Thats to say n is a factor of either 5 or m. If n is a factor of 5. The only value it can be is 5, but that will lead to an incorrect equation: m = 1-m. But if its a factor of m. Then the equation becomes: m = 1 - (5m / km) where k is an integer coefficient. Therefore: (m - 1)km = - 5m; km² + km - 5m = 0; km² + (k-5)m = 0. Use quadratic formumar to find m. m = ((k-m)² +- √(k-m))/2k; This means that k-m is supposed to be a perfect square since any other value will make √(k-m) a decimal, thus the value of m also a decimal. (Will edit when am back)

[williamperez-hernandez]

Also, equation simplifies to 5m=n(1-m). Assume m, n positive integers, then LHS is >=5 (5 when m=1). But RHS is <= 0 (zero for m=1). This is a contradiction. So m, n cannot both be positive integers.

[joyli]

When n gets closer to infinity, m gets closer to one.

[MrGeorge]

Assuming n≠-5 multiplying both side by n+5 gives mn+5m=n. Factorizing the equation: (1-m)(5+n)=5. 5 is prime so there are four cases for the two factors on LHS: (+1)∙(+5), (-1)∙(-5), (+5)∙(+1) and (-5)∙(-1).

So (OK, ignoring the positive integers restriction) we also have four solutions: (m, n) = (0, 0), (2, -10), (-4, -4) and (6, -6).

[MichaelRothwell]

Since n+5>n>0, we cannot have n+5|n. Done.

[speakingsarcasm]

As n tends to infinity, m tends to 1.... That's not a solution though...!

[Caloteira]

Sou Brasileiro não entendi o enunciado ingles ;-;