Finding an irreducible polynomial over the integers. I am wondering if we can find an irreducible polynomial $g(x)$ in $\mathbb{Z}[x]$ such that 


*

*The constant term, $c(g)=\pm 1$ and the leading coefficient $\ell(g)=\pm 1$, 

*the ideal generated by $g(x)$ and $5x+7$ is the ring $\mathbb{Z}[x]$, that is,   $(g(x),5x+7)=1$, in other words: $g(-7/5)=\pm 1$,

*the ideal generated by $g(x)$ and $2x-3$ is the ring $\mathbb{Z}[x]$, that is, 
$(g(x), 2x-3)=1$, in other words: $g(3/2)=\pm 1$.


Thanks.
PS: You can change $5x+7$ and $2x-3$ with any polynomials such that their constant terms and the leading coefficients are not units in $\mathbb{Z}$. 
 A: $g(x)=x^2-2$ works. (I'm taking $c$ to be the content, and $\ell$ to be the leading coefficient). Verification let to the reader, in case this is homework. 
EDITed in the light of comments and edits to the question statement: 
If non-constant $g$ has integer coefficients and leading coefficient $\pm1$, then $g(3/2)=\pm1$ is obviously impossible. However, $g(3/2)=\pm1$ has nothing to do with the ideal generated by $g$ and $2x-3$. 
For example, if $g(x)=x^2-x-1$, then $g$ satisfies all your conditions: it is irreducible, has integer coefficients, has leading coefficient and constant term $\pm1$, and with $2x-3$ generates the ring ${\bf Z}[x]$, as is evident from $$(-4)(x^2-x-1)+(2x+1)(2x-3)=1$$
I don't know whether there is a polynomial that satisfies your conditions simultaneously for $2x-3$ and $5x+7$, but if I find one, I'll let you know. 
Further EDIT: You say I can change the polynomials. If I change the $5x+7$ to $5x-8$ then $x^2-x-1$ will solve your problem. 
Even more EDIT: I'm now confident (but not 100% certain, since I haven't carried out all the calculations) that there is a polynomial $g$ of degree 6 with integer coefficients, leading coefficient 1, constant term $-1$, irreducible over the rationals, such that $1$ is in both the ideals $(g,5x+7)$ and $(g,2x-3)$. 
The condition on the ideals will be satisfied if $5^6g(-7/5)=-1$ and $2^6g(3/2)=-1$. Let $$g(x)=x^6+ax^5+bx^4+cx^3+dx^2+ex-1$$ Then we get the two equations in $5$ unknowns, $$7^6-7^55a+7^45^2b-7^35^3c+7^25^4d-(7)5^5e-5^6=-1$$ and $$3^6+3^52a+3^42^2b+3^32^3c+3^22^4d+(3)2^5e-2^6=-1$$ Move the constant terms to the right side of the equations, divide the 1st one by $-35$ (note that  $7^6-5^6+1$ is a multiple of $35$) and the second one by $6$ ($3^6-2^6+1$ is a multiple of $6$), and you have two linear equations in $5$ unknowns, with no modular obstacle to a solution. 
Now I wave my hands a little and say there must be infinitely many integer solutions to this pair of equations, including infinitely many with $g$ irreducible. In any event, it should not be hard to find one such solution. 
Please get back to me if there are any questions about this.   
A: Hint $\rm\ \ (g(x),\:3-2x)\: =\: 1\:$ in $\rm\:\mathbb Z[x]\iff 3-2x\ |\ 1\:$ in $\rm\: R\: =\: \mathbb Z[x]/g(x).\:$
If $\rm\:g(x)\: =\: x^2 - d\:$ then $\rm\:R = \mathbb Z[\sqrt{d}]\:$ so $\rm\:3-2\sqrt{d}\ |\ 1\:\Rightarrow\: 9-4d\: =\: \pm1\:$ by taking norms.
Similarly $\rm\:7+5\sqrt{d}\ |\ 1\:\Rightarrow\: 49-25d\ =\: \pm 1.$
So it suffices to solve $\rm\: 9-4d\: =\: \pm1,\ \ 49-25d\: =\: \pm 1,\:$ yielding $\rm\:d = 2,\ \ g\: =\: x^2 -2.$
