Prove the difference of any 2 squares of this sequence is a multiple of 24

[HelpMeMyTeacherHatesMe]

7, 13, 19, 25, 31
Prove that the difference of any two squares of this sequence is a multiple of 24.

I have been doing this and i have got it as far as
difference= 12(3n²+n-3m²-m)
but i cant get 24 on the outside of the bracket and still prove its an integer on the inside of the bracket

Thanks in advance

 

[Harry_the_cat]

7, 13, 19, 25, 31
Prove that the difference of any two squares of this sequence is a multiple of 24.

I have been doing this and i have got it as far as
difference= 12(3n²+n-3m²-m)
but i cant get 24 on the outside of the bracket and still prove its an integer on the inside of the bracket

Thanks in advance

12(3n²+n-3m²-m) ..... where n and m are integers
=12(3n²- 3m²+ n - m)
=12[3(n²-m²) +1(n-m)]
=12[3(n-m)(n+m)+1(n-m)]
=12(n-m)[3(n+m)+1]

So 12 is a factor.

Now looking at the other factors ie (n-m) and [3(n+m)+1]


If n and m are both even, (n-m) is even and so 12*2=24 is a factor of 12(n-m)[3(n+m)+1].


If n and m are both odd, (n-m) is even and so 12*2=24 is a factor of 12(n-m)[3(n+m)+1].


If m is even and n is odd (or vice versa), (m-n) will be odd; BUT (n+m) will be odd, and 3(n+m) will be odd and so [3(n+m)+1] will be even, and so 12*2 = 24 will be a factor of 12(n-m)[3(n+m)+1].

So in all possible cases 24 is a factor.

 

[HallsofIvy]

Have you left something out? There are only 5 numbers in the sequence you give so only 10 pairs of squares. You can calculate each one separately.

Are we to assume that this is the arithmetic sequence with first term 7 and common difference 6?

That is, \(\displaystyle a_n= 7+ 6n\)?

 

[HelpMeMyTeacherHatesMe]

Have you left something out? There are only 5 numbers in the sequence you give so only 10 pairs of squares. You can calculate each one separately.

Are we to assume that this is the arithmetic sequence with first term 7 and common difference 6?

That is, \(\displaystyle a_n= 7+ 6n\)?

Ahh yes I should of said. It is for any number in the sequence 6n+1

 

[HelpMeMyTeacherHatesMe]

.

12(3n²+n-3m²-m) ..... where n and m are integers
=12(3n²- 3m²+ n - m)
=12[3(n²-m²) +1(n-m)]
=12[3(n-m)(n+m)+1(n-m)]
=12(n-m)[3(n+m)+1]

.

Thank you so much, this really helped. However could you elaborate on how
12[3(n-m)(n+m)+1(n-m)]=12(n-m)[3(n+m)+1]

 

[Deleted member 4993]

Thank you so much, this really helped. However could you elaborate on how
12[3(n-m)(n+m)+1(n-m)]=12(n-m)[3(n+m)+1]

Do you see that (n-m) is a common factor in the left-hand-side of the equation?

 

[HelpMeMyTeacherHatesMe]

Do you see that (n-m) is a common factor in the left-hand-side of the equation?

So they just cancel out? Ok thanks for clearing that up

 

[Harry_the_cat]

Thank you so much, this really helped. However could you elaborate on how
12[3(n-m)(n+m)+1(n-m)]=12(n-m)[3(n+m)+1]

By the way, I'm sure your teacher doesn't hate you!!:-D

 

[Deleted member 4993]

So they just cancel out? Ok thanks for clearing that up

NO.....

It is simply factored out in front!!

 

[lookagain]

7, 13, 19, 25, 31
Prove that the difference of any two squares of this sequence is a multiple of 24.

The problem statement should be this:

"Prove that the difference of any two squares of the terms of this sequence is a multiple of 24."