Notations:
$R$- Noetherian graded ring and $I,J$ homogeneous ideals in $R$


**Definition:**
 The projective dimension of $R/I$, denoted $pd(R/I)$, is the length of a minimal free graded resolution of $R/I$:
 $0 \rightarrow \bigoplus\_{j} R(-j)^ {\beta\_{p,j}(R/I)}\rightarrow \bigoplus\_{j} R(-j)^ {\beta\_{p-1,j}(R/I)} \rightarrow \cdots \rightarrow \bigoplus\_{j} R(-j)^ {\beta\_{0,j}(R/I)} \rightarrow 
 R/I \rightarrow 0$.


where $p \leq n$ and $R(-j)$ indicates the ring $R$ with the shifted graduation such that, for all $a \in \mathbb{Z}$, $R(-j)\_{a}=R\_{a-j}$ and
$\beta\_{i,j}(R/I)$ is the number of copies of $R(-j)$ appearing in the $i$-th module of the resolution, 
and is called graded Betti number degree $(i,j)$. The $i$-th Betti number is $\beta\_{i}(R/I)=\sum \limits\_{j \in \mathbb{Z}}\beta\_{i,j}(R/I)$. 
The Castelnuovo-Mumford regularity (or simply regularity) of $R/I$ is



> 
> $\operatorname{reg}(R/I)=\max\{j-i\mid\beta\_{i,j}(R/I)\neq 0\}$
> 
> 
> 


Let $I$ be quadratic square free monomial ideal in $k[x\_1,\ldots,x\_n]$ and $J$ be quadratic square free monomial ideal in $k[y\_1,\ldots,y\_m]$ with $I\ncong J$. Suppose $\beta\_{i,2i}(I) \neq 0$ and $\beta\_{j,2j}(J) \neq 0$ for some $i,j$. Let $IJ$ be square free monomial in $k[x\_1,\ldots,x\_n,y\_1,\ldots,y\_m]$.



> 
> Can we say that $\beta\_{(i+j+2),2(i+j+2)}(IJ)\neq 0$ ?
> 
> 
> 



Properties of Betti number of ideal
I believe the answer to your question is no, we cannot conclude that $\beta\_{(i+j+2), 2(i+j+2)}(IJ)\neq0$.


For example, let us take $n=m=4$ and $I$ and $J$ to be the following ideals
$$I = \langle x\_1x\_2, x\_2x\_3\rangle \subset \mathbb{Q}[x\_1, x\_2, x\_3, x\_4]=R$$
$$J = \langle y\_1y\_2, y\_3y\_4\rangle \subset \mathbb{Q}[y\_1, y\_2, y\_3, y\_4]=S.$$
Both of these are generated by square-free monomials and by considering dimensions we see $R/I$ is not isomorphic to $S/J$. (I assume this is what you mean by $I\not\cong J$, sorry if I missed interpreted.) Using Macualay2 to compute the Betti tables of $I$, $J$, and $IJ$ we get:


$$b(I)=
\begin{array}{c | c c c c c}
- & 0 & 1 & 2 \\ \hline
T & 1 & 2 & 1 \\
0 & 1 & - & - \\
1 & - & 2 & 1 \\
\end{array}
\quad 
b(J)=
\begin{array}{c | c c c c c}
- & 0 & 1 & 2 \\ \hline
T & 1 & 2 & 1 \\
0 & 1 & - & -\\
1 & - & 2 & - \\
2 & - & - & 1
\end{array}
\quad
b(IJ)=
\begin{array}{c | c c c c c}
- & 0 & 1 & 2 & 3 \\ \hline
T & 1 & 4 & 4 & 1 \\
0 & 1 & - & - & - \\
1 & - & - & - & - \\
2 & - & - & - & - \\ 
3 & - & 4 & 2 & - \\
4 & - & - & 2 & 1 
\end{array}$$


In particular, we see that: 
 $$b\_{1,1}(I)=\beta\_{1,2}(I)=2, \quad \quad b\_{1,1}(J)=\beta\_{1,2}(J)=2, \quad \quad \text{and} \quad\quad b\_{2,2}(J)=\beta\_{2,4}(J)=1.$$
So we may take $i=1$ and $j=1,2$, but in either case:
$$\beta\_{1+1+2, 2(1+1+2)}(IJ)=b\_{4,4}(IJ)=0 \quad \quad \text{and} \quad \quad \beta\_{1+2+2, 2(1+2+2)}(IJ)=b\_{5,5}(IJ)=0.$$
Out of curiosity what lead you to believe this might be true? (I don't have a lot of intuition for this, and found the above by just messing mourned in Macaulay2.) 


