(8+1)^2 = 81

[apple2357]

I am trying to establish whether there are other numbers ( two digits for the moment!) that have the curious property (8+1)^2 = 81

Using a bit of algebra ( k+1)^2= 10k +1, it can be established that k=8 is the only solution ( ignoring k=0), but this is a particular case with a unit digit of 1.

Generalising further:

(k+m)^2 = 10k + m

And here i get stuck, how do i establish solutions to this equation other than the above: k=8, m=1 ?

In other words, can anyone offer me any strategies to prove that there is only one solution ( if that is indeed the case?) to

k^2 + 2km + m^2 -10k-m =0

Can i use things like discriminant? Doesn't quite make sense with two variables...

 

[Harry_the_cat]

Something to consider: If 10k+m is a perfect square then m must be chosen from 1, 4, 9, 5 or 6.

Also if you are only considering 2 digit numbers for now, the you can test out all of the 2-digit squares (there are only 6 of them).

 

[Deleted member 4993]

I am trying to establish whether there are other numbers ( two digits for the moment!) that have the curious property (8+1)^2 = 81

Using a bit of algebra ( k+1)^2= 10k +1, it can be established that k=8 is the only solution ( ignoring k=0), but this is a particular case with a unit digit of 1.

Generalising further:

(k+m)^2 = 10k + m

And here i get stuck, how do i establish solutions to this equation other than the above: k=8, m=1 ?

In other words, can anyone offer me any strategies to prove that there is only one solution ( if that is indeed the case?) to

k^2 + 2km + m^2 -10k-m =0

Can i use things like discriminant? Doesn't quite make sense with two variables...

I think investigating the discriminant will be very helpful!!