<s>[INST] Divisibility question Prove:  
(A) sum of two squares of two odd integers cannot be a perfect square 
(B) the product of four consecutive integers is $1$ less than a perfect square 

For (A) I let the two odd integers be $2a + 1$ and $2b + 1$ for any integers $a$ and $b$. After completing the expansion for sum of their squares , I could not establish the link to it not being a perfect t square 
For (B) I got stuck at expanding $(a)(a + 1)(a+2) (a + 3)$
Could someone help please ?
Thanks 
 [/INST] A)
\begin{align*}
&\,(2a+1)^2+(2b+1)^2=(4a^2+4a+1)+(4b^2+4b+1)=4(a^2+b^2)+4(a+b)+2\\
=&\,2[2(a^2+b^2)+2(a+b)+1].
\end{align*}
Since $2(a^2+b^2)$ and $2(a+b)$ are both even, the expression between the brackets is odd because of the $+1$ term. Now, the double of an odd number can never be a perfect square. (Try proving this last statement.)
B)
\begin{align*}
&\,a(a+1)(a+2)(a+3)=a(a+3)[(a+1)(a+2)]=a(a+3)[a^2+3a+2]\\
=&\,a(a+3)[a(a+3)+2]=a^2(a+3)^2+2a(a+3)=[a^2(a+3)^2+2a(a+3)+1]-1\\
=&\,[a(a+3)+1]^2-1.
\end{align*}
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