(a) 25=5^2. If the number is not divisible by 5, it is not also divisible by 25.
Let's see if n^2+n-9 is divisible by 5 or not.
For example, if n is divisible by 5, there is an integer k that satisfies n=5k.
Substitute n=5k into n^2+n-9:
n^2+n-9=(5k)^2+(5k)-9
=25k^2+5k-9
=5(5k^2+k-2)+1.
Thus, the remainder is 1. n^2+n-9 is not divisible by 5 if n=5k.
We can express this in congruence.
If n≡0 (mod 5) then n^2+n-9≡0+0-9≡-9≡1 (mod 5)
For other cases(n≡1,2,3,4 (mod 5)), see the table.
![enter image source here]()
n^2+n-9 is not a multiple of 5 in any cases, neither is it a multiple of 25.
(b) 49=7^2.
Similarly to (a), find the remainder (mod 7) first.
![enter image source here]()
If n≡0,2,3,4,6 (mod 7), n^2+n-9 is a not divisible by 7 and it is proven that n^2+n-9 is not a multiple of 49.
If n≡1 or n≡5 (mod 7), n^2+n-9 is a multiple of 7.
Now let's move to the next stage.
How to show that no natural square n=m^2 satisfies (A),(B)?
(A) n≡1,5 (mod 7)
(B) n^2+n-9≡0 (mod 49)
Since m^2≡0,1,2,4 (mod 7), we can confine to the case n=m^2≡1 (mod 7). In this case, m must satisfy m≡1,6 (mod 7) , i.e. m≡1,6,8,13,15,20,22,27,29,34,36,41,43,48 (mod 49).
Here is a table for m, n=m^2, n^2=m^4 and n^2+n-9=m^4+m^2-9.
![enter image source here]()
Oops! n^2+n-9=m^4+m^2-9 is divisible by 49 if m≡6,43 (mod 49) and the statement is false.
For example:
m=6, n=6^2=36
n^2+n-9=36^2+36-9=1323=49*27.
