How would I start off this question?

[EddyBenzen122]

for some real numbers and n, the list 3n2 m2 2(n + 1)2 consists of three consecutive integers written in increasing order. determine all possible values of m.

I have no idea how I can start this equation and so I couldn't show any work. Would you mind helping me by giving me hints or whatever?

 

[skeeter]

did you mean real numbers m and n, with the sequence

[imath]3n^2, \, m^2, \, 2(n+1)^2[/imath] ???

if so, note the last term, [imath]2(n+1)^2[/imath], is an even number (why?)

see what you can do from here

 

[BigBeachBanana]

for some real numbers and n, the list 3n2 m2 2(n + 1)2 consists of three consecutive integers written in increasing order. determine all possible values of m.

I have no idea how I can start this equation and so I couldn't show any work. Would you mind helping me by giving me hints or whatever?

You were told that they are increasing consecutive integers so:
[math]\tag{eq 1} m^2-3n^2=1[/math][math]\tag{eq 2} 2(n+1)^2-m^2=1[/math]It's a system of 2 equations and 2 unknowns. Solve for [imath]m,\,n[/imath] simultaneously.

 

[lex]

[imath]2(n+1)^2-3n^2=2[/imath]

(Finally, note there are 4 answers for m).

 

[skeeter]

[imath]2(n+1)^2-3n^2=2[/imath]

(Finally, note there are 4 answers for m).

two solutions for n … how does that yield four solutions for m?

 

[Steven G]

two solutions for n … how does that yield four solutions for m?

Suppose for a moment that n=1 and n=2. Then m^2 =4 or m^2 = 13. Mathematically that yields 4 values for m.

 

[skeeter]

Suppose for a moment that n=1 and n=2. Then m^2 =4 or m^2 = 13. Mathematically that yields 4 values for m.

ok … I was just looking at possibilities for m^2