HOW TO PROVE THAT :  49+56(N^2 +1)  CAN NEVER BE …

Let’s simplify the expression:

\[
49 + 56(n^2 + 1) = 49 + 56n^2 + 56 = 56n^2 + 105
\]

We need to prove that:

\[
k^2 \ne 56n^2 + 105 \quad \text{for any integer } k \text{ and } n \in \mathbb{N}_0
\]

Proof by Contradiction:

Assume there exists some \( n \in \mathbb{N}_0 \) and \( k \in \mathbb{Z} \) such that:

\[
k^2 = 56n^2 + 105
\]

Rewriting:

\[
k^2 – 56n^2 = 105
\]

This is a Diophantine equation of the form:

\[
k^2 – 56n^2 = 105
\]

It resembles a generalized Pell’s equation, but unlike standard Pell’s equations, this has a non-zero right-hand side.

To find integer solutions, test small values of \( n \):

– \( n = 0 \Rightarrow k^2 = 105 \) → not a perfect square
– \( n = 1 \Rightarrow k^2 = 56 + 105 = 161 \) → not a perfect square
– \( n = 2 \Rightarrow k^2 = 224 + 105 = 329 \) → not a perfect square
– \( n = 3 \Rightarrow k^2 = 504 + 105 = 609 \) → not a perfect square
– \( n = 4 \Rightarrow k^2 = 896 + 105 = 1001 \) → not a perfect square
– \( n = 5 \Rightarrow k^2 = 1400 + 105 = 1505 \) → not a perfect square
– \( n = 6 \Rightarrow k^2 = 2016 + 105 = 2121 \) → not a perfect square

And so on. No value of \( k^2 = 56n^2 + 105 \) becomes a perfect square for any non-negative integer \( n \).

Also note:
For \( k^2 \equiv 56n^2 + 105 \pmod{8} \), since:

\[
56n^2 \equiv 0 \pmod{8}, \quad \Rightarrow k^2 \equiv 105 \equiv 1 \pmod{8}
\]

Only \( k \equiv 1, 3, 5, 7 \pmod{8} \) will work. However, checking modulo 7:

\[
56n^2 + 105 \equiv 0n^2 + 0 = 0 \pmod{7}
\Rightarrow k^2 \equiv 0 \pmod{7}
\Rightarrow k \equiv 0 \pmod{7}
\]

So \( k = 7m \), and we get:

\[
(7m)^2 = 56n^2 + 105 \Rightarrow 49m^2 = 56n^2 + 105
\Rightarrow 7m^2 = 8n^2 + 15
\]

Now check modulo 7:

\[
8n^2 + 15 \equiv m^2 \pmod{7}
\Rightarrow (8n^2 + 15) \mod 7
\]

But trying all \( n = 0 \) to \( 6 \), none of the RHS becomes a multiple of 7 ⇒ contradiction.

Conclusion:

\[
\boxed{49 + 56(n^2 + 1) \text{ is never a perfect square for any } n \in \mathbb{N}_0}
\]