# When is $\displaystyle \prod_i \prod_j a_{i} a_{j} = \Bigl(\prod_i a_i\Bigr)^2$

In statistical mechanics, I used to use the procedure that if $a_{ij}=a_i a_j$ $$\prod_i\; \prod_j a_{i}a_{j} = \biggl(\prod_i a_i\biggr)\vphantom{\Bigr)}^2$$

However, today I noticed, $$\prod_i\; \prod_j 2^{i+j} = 2^{\sum_i\sum_j(i+j)}=2^{n^2(n+1)}$$

$$\prod_i\; \prod_j 2^{i+j} =\prod_i 2^i \; \prod_j 2^j = \biggl(\prod_i2^i\biggr)\vphantom{\Bigr)}^2 = 2^{n(n+1)}$$

Why does the second method fail. When is it applicable?

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It feels that there was some confusion about sum and product. $$\sum_{i=1}^n \sum_{j=1}^n a_ia_j= (\sum_{i=1}^n a_i)^2.$$ –  Jiangwei Xue Jun 25 '11 at 19:30
That identity is just wrong. Replace the products by summations, or raise the right hand side to the nth power, and you're good. –  George Lowther Jun 25 '11 at 19:32
@George so $\prod \prod a_i a_j =\left(\prod_ia_i\right)^{2n}$? –  kuch nahi Jun 25 '11 at 19:35
@Jiangwei true, thanks. I realize my confusion now –  kuch nahi Jun 25 '11 at 19:37
@George I edited it, thanks. Could you post it as an answer. –  kuch nahi Jun 25 '11 at 19:39

The identity in the question is wrong. The correct one is $$\prod_{i=1}^n\prod_{j=1}^na_ia_j=\left(\prod_{i=1}^na_i\right)^{2n}.$$ As a quick `sanity check' you can try counting the number of a's in the product on each side. The expression in the question had $2n^2$ on the left but only $2n$ on the right, so couldn't possibly be correct. One possible source of confusion is that the original expression would be correct if you had summation signs rather than products, $$\sum_{i=1}^n\sum_{j=1}^na_ia_j=\left(\sum_{i=1}^na_i\right)^2.$$