(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.

#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.

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#.

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#.