This is from the polynomial example $x^4 + 3 x^2 + 7x + 4$ discussed in the framework of Chebotarev Density in Lenstra-Stevenhagen, especially pages 10,11,12 in this pdf. The article appeared in 1996 in The Mathematical Intelligencer.
I guess I should repeat that they point out that $x^4 + 3 x^2 + 7x + 4$ is reducible mod every prime. This has thrown me off, I had this idea that reducibility would mean that the homogeneous polynomial below, the norm form, would integrally represent low powers of that prime, especially $p$ or $p^2$ or $p^3.$ Something very much like that happened with these earlier questions,
what numbers are integrally represented by this quartic polynomial (norm form)
primes represented integrally by a homogeneous cubic form
I thought I had something reasonable which is to take the companion matrix $$ A = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -4 & -7 & -3 & 0 \end{array} \right) $$ and create $$ f(x,y,z,t) = \det \left( x I + y A + z A^2 + t A^3 \right) $$ $$ f(x,y,z,t) = x^4 + (-6z - 21t)x^3 + (3y^2 + (21z - 2t)y + (17z^2 + 21tz + 138t^2))x^2 + (-7y^3 + (-16z + 42t)y^2 + (-21z^2 - 99tz - 203t^2)y + (25z^3 + 28tz^2 + 139t^2z - 91t^3))x + (4y^4 - 24ty^3 + (12z^2 + 84tz + 68t^2)y^2 + (-28z^3 - 64tz^2 - 84t^2z + 100t^3)y + (16z^4 + 48t^2z^2 - 112t^3z + 64t^4)) $$ Now, this is automatically completely multiplicative. If there is a quadruple of integers that cause $f$ to take the value $m,$ another quadruple that gives $n,$ we can explicitly find (using Cayley-Hamilton) a quadruple by which $f$ represents $mn.$
I had a fairly simple picture of what numbers would pop up, but nothing seems to work properly with this one. For one thing, it seems we cannot get quadratic non-residues $\pmod 7,$ which already strikes me as cheating.
Here are the first 150 represented primes:
jagy@phobeusjunior:~$ ./Lenstra_Chebotarev
1 29 53 71 137 149 163 193 263 317
337 421 499 541 547 557 599 613 617 631
641 673 701 709 739 743 751 809 823 1031
1061 1087 1103 1163 1229 1283 1327 1367 1373 1409
1423 1453 1471 1499 1579 1583 1619 1621 1733 1789
1933 1997 2017 2039 2069 2081 2087 2111 2137 2153
2179 2207 2221 2237 2251 2293 2297 2333 2347 2377
2389 2447 2473 2503 2521 2531 2543 2683 2689 2713
2797 2909 2969 3011 3187 3221 3259 3271 3301 3319
3371 3467 3511 3593 3623 3677 3691 3803 3833 3847
3889 3907 3917 4013 4019 4127 4139 4201 4211 4243
4271 4327 4337 4349 4391 4421 4463 4621 4649 4691
4729 4733 4817 4831 4937 4943 4957 4993 4999 5051
5153 5209 5231 5413 5419 5443 5483 5503 5527 5531
5573 5623 5653 5693 5779 5783 5791 5839 5849 5861
Hmmmm. It seems that when $p$ is not represented, we can represent $2 p^2$
18 = 2 3^2 mod seven 4 mod nineteen 18
50 = 2 5^2 mod seven 1 mod nineteen 12
98 = 2 7^2 mod seven 0 mod nineteen 3
242 = 2 11^2 mod seven 4 mod nineteen 14
338 = 2 13^2 mod seven 2 mod nineteen 15
578 = 2 17^2 mod seven 4 mod nineteen 8
722 = 2 19^2 mod seven 1 mod nineteen 0
2738 = 2 37^2 mod seven 1 mod nineteen 2
3362 = 2 41^2 mod seven 2 mod nineteen 18
4418 = 2 47^2 mod seven 1 mod nineteen 10
6962 = 2 59^2 mod seven 4 mod nineteen 8
7442 = 2 61^2 mod seven 1 mod nineteen 13
10658 = 2 73^2 mod seven 4 mod nineteen 18
15842 = 2 89^2 mod seven 1 mod nineteen 15
18818 = 2 97^2 mod seven 2 mod nineteen 8
20402 = 2 101^2 mod seven 4 mod nineteen 15
22898 = 2 107^2 mod seven 1 mod nineteen 3
25538 = 2 113^2 mod seven 2 mod nineteen 2
34322 = 2 131^2 mod seven 1 mod nineteen 8
38642 = 2 139^2 mod seven 2 mod nineteen 15
45602 = 2 151^2 mod seven 4 mod nineteen 2
49298 = 2 157^2 mod seven 4 mod nineteen 12
55778 = 2 167^2 mod seven 2 mod nineteen 13
59858 = 2 173^2 mod seven 1 mod nineteen 8
64082 = 2 179^2 mod seven 4 mod nineteen 14
65522 = 2 181^2 mod seven 2 mod nineteen 10
72962 = 2 191^2 mod seven 1 mod nineteen 2
77618 = 2 197^2 mod seven 2 mod nineteen 3
99458 = 2 223^2 mod seven 2 mod nineteen 12
On that note, when $p,q$ are primes that are not represented, we can represent $pq$ as long as $pq \equiv 0,1,2,4 \pmod 7.$
Here are the first 100 numbers represented with $\gcd(x,y,z,t) = 1.$
jagy@phobeusjunior:~$ ./Lenstra_Chebotarev | sort -n | head -100
1 = 1 mod seven 1 mod nineteen 1
4 = 2^2 mod seven 4 mod nineteen 4
15 = 3 5 mod seven 1 mod nineteen 15
16 = 2^4 mod seven 2 mod nineteen 16
18 = 2 3^2 mod seven 4 mod nineteen 18
21 = 3 7 mod seven 0 mod nineteen 2
29 = 29 mod seven 1 mod nineteen 10
46 = 2 23 mod seven 4 mod nineteen 8
50 = 2 5^2 mod seven 1 mod nineteen 12
53 = 53 mod seven 4 mod nineteen 15
60 = 2^2 3 5 mod seven 4 mod nineteen 3
64 = 2^6 mod seven 1 mod nineteen 7
65 = 5 13 mod seven 2 mod nineteen 8
70 = 2 5 7 mod seven 0 mod nineteen 13
71 = 71 mod seven 1 mod nineteen 14
72 = 2^3 3^2 mod seven 2 mod nineteen 15
78 = 2 3 13 mod seven 1 mod nineteen 2
81 = 3^4 mod seven 4 mod nineteen 5
84 = 2^2 3 7 mod seven 0 mod nineteen 8
85 = 5 17 mod seven 1 mod nineteen 9
86 = 2 43 mod seven 2 mod nineteen 10
91 = 7 13 mod seven 0 mod nineteen 15
95 = 5 19 mod seven 4 mod nineteen 0
98 = 2 7^2 mod seven 0 mod nineteen 3
102 = 2 3 17 mod seven 4 mod nineteen 7
114 = 2 3 19 mod seven 2 mod nineteen 0
116 = 2^2 29 mod seven 4 mod nineteen 2
119 = 7 17 mod seven 0 mod nineteen 5
133 = 7 19 mod seven 0 mod nineteen 0
134 = 2 67 mod seven 1 mod nineteen 1
137 = 137 mod seven 4 mod nineteen 4
149 = 149 mod seven 2 mod nineteen 16
158 = 2 79 mod seven 4 mod nineteen 6
163 = 163 mod seven 2 mod nineteen 11
177 = 3 59 mod seven 2 mod nineteen 6
183 = 3 61 mod seven 1 mod nineteen 12
184 = 2^3 23 mod seven 2 mod nineteen 13
193 = 193 mod seven 4 mod nineteen 3
200 = 2^3 5^2 mod seven 4 mod nineteen 10
205 = 5 41 mod seven 2 mod nineteen 15
207 = 3^2 23 mod seven 4 mod nineteen 17
212 = 2^2 53 mod seven 2 mod nineteen 3
218 = 2 109 mod seven 1 mod nineteen 9
219 = 3 73 mod seven 2 mod nineteen 10
225 = 3^2 5^2 mod seven 1 mod nineteen 16
235 = 5 47 mod seven 4 mod nineteen 7
240 = 2^4 3 5 mod seven 2 mod nineteen 12
242 = 2 11^2 mod seven 4 mod nineteen 14
246 = 2 3 41 mod seven 1 mod nineteen 18
254 = 2 127 mod seven 2 mod nineteen 7
256 = 2^8 mod seven 4 mod nineteen 9
260 = 2^2 5 13 mod seven 1 mod nineteen 13
263 = 263 mod seven 4 mod nineteen 16
267 = 3 89 mod seven 1 mod nineteen 1
270 = 2 3^3 5 mod seven 4 mod nineteen 4
280 = 2^3 5 7 mod seven 0 mod nineteen 14
282 = 2 3 47 mod seven 2 mod nineteen 16
284 = 2^2 71 mod seven 4 mod nineteen 18
287 = 7 41 mod seven 0 mod nineteen 2
288 = 2^5 3^2 mod seven 1 mod nineteen 3
303 = 3 101 mod seven 2 mod nineteen 18
312 = 2^3 3 13 mod seven 4 mod nineteen 8
315 = 3^2 5 7 mod seven 0 mod nineteen 11
317 = 317 mod seven 2 mod nineteen 13
324 = 2^2 3^4 mod seven 2 mod nineteen 1
329 = 7 47 mod seven 0 mod nineteen 6
336 = 2^4 3 7 mod seven 0 mod nineteen 13
337 = 337 mod seven 1 mod nineteen 14
338 = 2 13^2 mod seven 2 mod nineteen 15
340 = 2^2 5 17 mod seven 4 mod nineteen 17
344 = 2^3 43 mod seven 1 mod nineteen 2
351 = 3^3 13 mod seven 1 mod nineteen 9
364 = 2^2 7 13 mod seven 0 mod nineteen 3
378 = 2 3^3 7 mod seven 0 mod nineteen 17
380 = 2^2 5 19 mod seven 2 mod nineteen 0
387 = 3^2 43 mod seven 2 mod nineteen 7
392 = 2^3 7^2 mod seven 0 mod nineteen 12
393 = 3 131 mod seven 1 mod nineteen 13
408 = 2^3 3 17 mod seven 2 mod nineteen 9
417 = 3 139 mod seven 4 mod nineteen 18
421 = 421 mod seven 1 mod nineteen 3
422 = 2 211 mod seven 2 mod nineteen 4
435 = 3 5 29 mod seven 1 mod nineteen 17
441 = 3^2 7^2 mod seven 0 mod nineteen 4
442 = 2 13 17 mod seven 1 mod nineteen 5
456 = 2^3 3 19 mod seven 1 mod nineteen 0
459 = 3^3 17 mod seven 4 mod nineteen 3
464 = 2^4 29 mod seven 2 mod nineteen 8
466 = 2 233 mod seven 4 mod nineteen 10
476 = 2^2 7 17 mod seven 0 mod nineteen 1
485 = 5 97 mod seven 2 mod nineteen 10
494 = 2 13 19 mod seven 4 mod nineteen 0
499 = 499 mod seven 2 mod nineteen 5
501 = 3 167 mod seven 4 mod nineteen 7
513 = 3^3 19 mod seven 2 mod nineteen 0
519 = 3 173 mod seven 1 mod nineteen 6
522 = 2 3^2 29 mod seven 4 mod nineteen 9
529 = 23^2 mod seven 4 mod nineteen 16
532 = 2^2 7 19 mod seven 0 mod nineteen 0
536 = 2^3 67 mod seven 4 mod nineteen 4
jagy@phobeusjunior:~$
The fact you always get a square modulo 7 is easy to explain:
$$ x^4 + 3x^2 + 7x + 4 \equiv (x-3)^2 (x-4)^2 \pmod 7 $$
Since your form $f$ is given by taking norms, which in turn are given by resultants (I may have the ordering of the arguments wrong):
$$ f(a,b,c,d) = \operatorname{Res}_x(x^4 + 3x^2 + 7x + 4, a + bx + cx^2 + dx^3) $$
The fact the left hand side is a square modulo $7$ guarantees the resultant is square modulo $7$.