Not a Multiple of 72

The idea here when one sees the word prime is to factorise the given expression. This is because when we factorise, we can conclude that some of the products must be $1$ , which helps limit the cases and we can also use coprime arguments to further reduce the cases to check manually.

The other idea is to try small cases . This comes in handy in most questions when you do not know how to start. For our question, we put in $p = 2$ and we get $18$ and $p = 3$ gives $88$ . However, $p = 5$ and $p = 7$ gives $648$ and $2448$ respectively, both of which are divisible by $72$ . This will definitely motivate us to conjecture that $p = 2,3$ are the only solutions.

Now, turning back to idea 1, we factorise as follows, using the fact that $(±1)^4 + (±1)^2 - 2 = 0$ , we easily get:

$p^4 + p^2 - 2 = ( p + 1)(p-1)(p^2 + 2)$

As I mentioned, we would like to consider $p > 3$ . If we can show that the expressions contribute a total factor of $72 = 3^2 \times 8$ then we are done. Now clearly since $p$ is prime, $3 \nmid p$ . So one of $(p + 1), (p - 1)$ are divisible by 3, and $(p+1)(p-1)$ is divisible by 3. By a stroke of luck, we realise that $p^2 + 2 = (p+1)(p-1) + 3$ so it is also divisible by 3.

So we are only left with showing that $8$ divides the expression. This is simple if we notice that $p$ is obviously odd by definition, so we have two consecutive even numbers, $(p+1)$ and $(p-1)$ , one of which must be divisible by $4$ and the other by $2$ . So we are done.

In conclusion, the answer to this problem is $2 \times 3 = 6$ .

First, we note that the polynomial above factors as $(p^2+2)(p+1)(p-1)$ . Now we use the fact that primes can be expressed in the form $6n\pm 1$ for $p\neq 2,3$ .

Substituting this into the factored expression, we get that the product is $(36n^2\pm 12n+3)(6n)(6n\pm 2)$ . Now we can take out the common factors to obtain $36(12n^2\pm 4n+1)(n)(3n\pm 1)$ .

Clearly if $n \equiv 0 \pmod{2}$ , then we have a multiple of $72$ . If $n \equiv 1 \pmod{2}$ , then $3n\pm 1\equiv 3\pm 1 \equiv 0 \pmod{2}$ , so this is also a multiple of $72$ . Thus, we have shown that for all prime numbers $p>3$ , $p^4+p^2-2$ is a multiple of $72$ .

Thus it remains to check $2$ and $3$ :

$(2^2+2)(2+1)(2-1)=6\cdot3\cdot1=18$ , which is not a multiple of $72$ .

$(3^2+2)(3+1)(3-1)=11\cdot4\cdot2=88$ , which is not a multiple of $72$ .

Therefore, $2$ and $3$ are the only numbers that do not produce multiples of $72$ , so the answer is their product: $2\cdot3=\boxed{6}$ .

Firstly, if p = 2, 3, $p^4 + p^2 - 2$ is not a multiple of 72.

Let $p \neq 2, 3$ . Note that the square of an odd number is 1 modulo 8. Since all primes are odd except for 2, $p^4 + p^2 - 2 \equiv 1 + 1 - 2 \equiv 0 \pmod 8$ .

Also since $p \neq 3$ and the square of a number not divisible by 3 is 1 modulo 3, $p^2 \equiv 1, 4, 7 \pmod 9$ . Hence $p^4 + p^2 - 2 \equiv 1 + 1 - 2, 16 + 4 - 2, 49 + 7 - 2 \equiv 0, 18, 54 \equiv 0 \pmod 9$ . Hence for $p \neq 2, 3$ , $p^4 + p^2 - 2 \equiv 0 \pmod 9$ . Hence $p^4 + p^2 - 2 \equiv 0 \pmod 8 \cdot 9 = 72$ since 8 and 9 are coprime.

So the only primes p that are valid are 2 and 3 $\Rightarrow$ the product is 6.

72=8.9 Let x be a prime not equal to 2 or 3 Let f(x)= ${x}^4 + x^2 -2$ $x^4 + x^2 -2=(x^2-1)(x^2+2)$ If x is prime and not 2, then $x \equiv \pm 1 \pmod 8$

or $x \equiv \pm 3 \pmod 8$ In any case ${x}^2-1 \equiv \0 \pmod{8}$ so f(x) is a multiple of 8. If x is prime and not 3 then $x \equiv \pm 1 \pmod 9$ or $x \equiv \pm 2 \pmod 9$ or $x \equiv \pm 4 \pmod 9$ In all these cases $f(x) \equiv \pm 0 \pmod 9$ So, for all primes not equal to 2 or 3 f(x) is a multiple of 72. Also f(2)=18 and f(3)=96, so they are the only primes which satisfy the condition. So the answer is 6.

note: The blank box is supposed to be x.

First we note that p^4+p^2-2 is not divisible by 72 for p=2,3. We may write the polynomial as p^{4}-1+p^{2}-1=(p^2-1)(p^2+2). Odd squares are always congruent to 1 modulo 8, so we see that 8 divides p^{2}-1 for all odd primes. Similarly, modulo 3, the square a of prime p\neq 3 must be congruent to 1. This means that both p^2-1 and p^2+2 will be divisible by 3 for primes p\neq 3. Since 72=9\cdot 8, we see that p^4+p^2-2 will be divisible by 8 and 9 for all primes p\neq 2,3. It follows that the answer to the problem is 6.

If $p>3$ , $\ \ p=6k\pm 1$ for a proper integer $k$ . We have: $(6k\pm 1)^4+(6k\pm 1)^2-2\equiv$ $\equiv (\pm24k+1)+(36k^2 \pm12k+1)-2 \equiv$ $\equiv 36k(k\pm 1)\equiv 0 \pmod{72} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ So the only $p$ such that $p^4+p^2-2$ is not multiple of $72$ are $2$ and $3$ (for wich we have respectively $18$ and $88$ ), whence the answer is $6$ .

Let $p = 2$ , $2^{4} + 2^{2} - 2 = 18 = 18 (mod 72);$ Let $p = 3$ , $3^{4} + 3^{2} - 2 = 88 = 16 (mod 72)$ ; Let $p = 5$ , $5^{4} + 5^{2} - 2 = 648 = 0 (mod 72)$ ; Let $p = 7$ , $7^{4} + 7^{2} - 2 = 2448 = 0 (mod 72);$ Let $p = 11$ , $11^{4} + 11^{2} - 2 = 14760 = 0 (mod 72)$ , ...... Since the rest of the primes give the same remainder, there are only 2 solutions. Hence, $2 \times{3} = \boxed{6}$

We prove that for any prime $p \geq 5, p^4 + p^2 - 2$ is a multiple of 72
We have $p^4 + p^2 - 2 = (p^2 - 1) (p^2 + 2)$ .
$p \geq 5 \Rightarrow p>3 \Rightarrow p$ not a multiple of $3 \Rightarrow p = 3n \pm 1$
Morever, $p$ is odd $\Rightarrow 3n$ is even $\Rightarrow n$ is even $\Rightarrow n = 2k$ $\Rightarrow p = 6k \pm 1 \Rightarrow p^2 = 36k^2 \pm 12k + 1$ $\Rightarrow p^2 - 1 =36k^2 \pm 12k=24k^2 + 12k(k\pm1) \vdots 24$
And $p^2 + 2 = 36k^2 \pm 12k + 3 \vdots 3\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad \quad\quad\quad\quad \quad$ So $p^4 + p^2 -2\vdots 72$ for any prime $p\geq 5\quad\quad\quad \quad\quad\quad\quad\quad\quad$
Other hand, $2^4 + 2^2 - 2 = 18$ (is not a multiple of 72) and $3^4 + 3^2 - 2 = 88$ (is not a multiple of 72). Finally, $2*3=6$

$p^{4}+p^{2}-2$ will be less than 72 for $p=2$ , and $p^{4}+p^{2}-2=90$ for $p=3$ , so that gives us our first 2 primes. Beyond that, we must turn to factoring.

$p^{4}+p^{2}-2$ can be factored into $(p^{2}-1)(p^{2}+2)$ , which can be further factored into $(p+1)(p-1)(p^{2}+2)$ . Since $p$ is prime and greater than 2, it must be odd; therefore, $p+1$ and $p-1$ must both be even numbers and divisible by 2. Further, either $p+1$ or $p-1$ must be divisible by 4 since every alternating even number is a multiple of 4. Also, $p-1$ , $p$ and $p+1$ are consecutive integers. One in every 3 consecutive integers is divisible by 3. Since $p$ is prime and greater than 3, either $p-1$ or $p+1$ must be divisible by 3. Thus, for $p+1$ and $p-1$ for every prime, one is divisible by 2 and the other is divisible by 4, and one is divisible by 3, so $p^{2}-1$ is divisible by $2^{3}3^{1}$ .

We can now examine $p^{2}+2$ . If $\frac{p^{2}+2}{3}$ is an integer, then it is divisible by 3. Let's rewrite $(p^{2}+2)$ as $(p^{2}-4+6)$ or $(p+2)(p-2)+6$ , which when plugged into our fraction becomes $\frac{(p+2)(p-2)+6}{3}=\frac{(p+2)(p-2)}{3}+\frac{6}{3}$ . Since we know $\frac{6}{3}$ is an integer, we can focus on determining if $\frac{(p+2)(p-2)}{3}$ is also an integer. $p-2$ , $p$ , and $p+2$ are consecutive odd integers. One of every 3 consecutive odd integers is divisible by 3. Again, since $p$ is prime and greater than 3, either $p-2$ or $p+2$ must be divisible by 3; therefore, for every prime, $p^{2}+2$ is divisible by 3.

So, for all remaining primes, $p^{4}+p^{2}-2$ is divisible by $2^{3}3^{2}$ , leaving us only 2 and 3 as our primes where $p^{4}+p^{2}-2$ is not a multiple of 72. $2 \times 3=6$

We try to factorize $f(p) = p^4 + p^2 - 2$ into factors with integer coefficients. This can be done by solving the equation $p^4 + p^2 - 2 = 0$ and multiply the two factors with the complex conjugate roots into a real polynomial. It is a second-degree equation in $p^2$ , and we get $p^2 = 1$ and $p^2 = -2$ . The four roots are 1, -1, $i \sqrt{2}$ and $-i \sqrt{2}$ . Thus $f(p) = (p-1)(p+1)(p-i \sqrt{2})(p+i \sqrt{2}) =$ $= (p-1)(p+1)(p^2+2)$ .

Assume p is a prime larger than 3 (and therefore odd).

For odd numbers p = 2m+1, both p-1 and p+1 are even. They are two consecutive even numbers, and thus one of them can be written either 4n and 4n+2, or 4n+2 and 4n+4, so one of them is divisible by 4. Their product (p-1)(p+1) is therefore divisible by 8.

Of the consecutive integers p-1,p,p+1, one must be divisible by 3. Since p is not divisible by 3, one of p-1 and p+1 must be, so (p-1)(p+1) is. Now looking at $p^2 +2$ , it can be written as $p^2 - 1 + 3 = (p-1)(p+1) + 3$ , and thus is divisible by 3 whenever the product (p-1)(p+1) is. We have therefore one factor 3 in (p-1)(p+1) and one in $p^2+2$ , making f(p) divisible by 9.

f(p) is divisible by both 8 and 9, which are coprime, so f(p) is divisible by their product 72 for all primes larger than 3.

The only thing left to test is the value of f(p) for the primes 2 and 3: f(2) = 18. f(3) = 88. Neither of those numbers is divisible by 72, so the product of the primes asked for is $2 \times 3 = 6$ .

p^4+p^2-2=(p^2+2)(p^2-1)= (p^2+2)(p-1)(p+1) if p != 2, then let p=2k+1, k=1,2,3……., then (p-1)(p+1)=4k(k+1), which can be divided by 8; if p=3k+1, k=1,2,3,….., then (p^2+2)(p-1)(p+1)=(9k^2+6k+3)(3k)(3k+2), which can be divided by 9; if p=3k-1, k=1,2,3,….., then (p^2+2)(p-1)(p+1)=(9k^2-6k+3)(3k)(3k-2), which can be divided by 9; So, we only leave the number 2 and 3, substitute to p^4+p^2-2, get the value not divided by 72 the answer is 6.

We immediately note that $p=2,3$ gives 18 and 88, respectively, which are not divisible by 72. We now claim that for all primes $p>3$ , we have that $p^4 + p^2 -2$ is divisible by 72. Note that any such prime can be written as $3k \pm 1$ , since it cannot be divisible by 3. Then, we remark that if $(p^2+23)(p^2 -22) = p^4 + p - 506 = p^4 + p^2 - 2 - 72 \cdot 7$ is divisible by 72, then clearly so must be $p^4 -p^2 -2$ . Then, $((3k\pm 1)^2 + 23)((3k\pm 1)^2 -22) = 9(3k^2 \pm 2k+ 8)(3k^2 \pm 2k - 7)$ . Hence, if $(3k^2 \pm 2k+ 8)(3k^2 \pm 2k - 7)$ is divisible by 8, then our number is divisible by 72. If $k$ is 0 mod 4, then clearly the first term is divisible by 8. If $k$ is 2 mod 4 (e.g. either 2 mod 8 or 6 mod 8), then the first term is 0 mod 8 (just substitute $k=2,6$ and algebra-bash). Finally, $k$ can't be 1 mod 4 or 3 mod 4, because this would give $p>3$ even, which is obviously impossible. Hence, $p^4+p^2 -2$ is a multiple of 72 for all primes $p>3$ , and our desired product is $2\cdot 3=6$ .

Let's say $p^4+p^2-2=Q(p)$ . Plugging the lowest primes into the function, we can see that Q(2)=18 and Q(3)=88 are not divisible by 72. Now we have to check larger primes. Any Q(p) divisible by 8 and 9 is divisible by 72. In other words, we can see whether any p is divisible by 72 by checking that $Q(p) \equiv 0 \pmod{8}, Q(p) \equiv 0 \pmod{9}$ .

Note: Q(2) is divisible by 9 but not 8, and Q(3) is divisible by 8 but not 9. We exclude 2, 4, 6 and 8 from our mod 8 calculations because multiples of 2 are not prime. The same reasoning applies to 3 and 6 in mod 9. This is why we dealt with 2 and 3 individually.

We only have to test 1, 3, 5 and 7 in mod 8. Because we're in mod 8, think of these four numbers as 1, 3, -3 and -1. Because Q(p) is an even function, we only have to plug in 1 and 3 and we find that $Q(\pm 1) \equiv Q(\pm 3) \equiv 0 \pmod{8}$ .

Using the same method, we have to test divisibility by 9. Excluding 3 and 6 because they are not prime, we have to test 1, 2, 4, 5, 7 and 8 mod 9, which can be rewritten as $\pm 1, \pm 2, \pm 4$ . For all these numbers $Q(p) \equiv 0 \pmod{9}$ , so we know that all primes greater than 3 will be divisible by 8 and 9, and therefore by 72.

Therefore, the only primes for which Q(p) is not a multiple of 72 are 2 and 3, and our answer is $2 \times 3=6$ .

Now if $p=2,3$ Then $p^4+p^2-2=18,88$ respectively, which are not multiples of 72.

If $p>3$ we note that all such primes are odd primes. So, $p^2 \equiv 1 \pmod{8}$ So, as $p^4+p^2-2=p^2(p^2+1)-2$ $p^2(p^2+1)-2 \equiv 1(1+1)-2=0 \pmod{8}$ So $p^4+p^2-2$ would be a multiple of 8.

Now, $p^2 \equiv 1,4,7 \pmod{9}$ And if we calculate $p^2(p^2+1)-2$ mod9, we will get; $p^2(p^2+1)-2 \equiv 0 \pmod{9}$ for each $p^2 \equiv 1,4,7 \pmod{9}$ Hence $p^4+p^2-2$ would also be a multiple of 9.

Therefore, $p^4+p^2-2$ would be a multiple of 72.

So the only $p$ values that will get produce a multiple of 72 is 2 and 3, so our required answer is $2*3=6$

Note that $2^4+2^2-2 = 18$ and $3^4+3^2-2 = 88$ which are clearly not divisible by 72.

Consider a prime $p>3$ . We shall show separately that $p^4+p^2-2 = p^2(p^2+1)-2$ is divisible by 9 and 8 for all such primes $p$

Let $p = 3k \pm 1$ for some integer $k$ since $p>3 \Rightarrow 3\nmid p$ . $p^2(p^2+1)-2=(3k\pm1)^2((3k\pm1)^2+1)-2$ $=(3(3k^2\pm2k)+1)(3(3k^2\pm2k)+2)-2$ $=9((3k^2\pm2k)^2+(3k^2\pm2k))$ which is divisible by 9 since $k$ is an integer.

Similarly, let $p = 4k \pm 1$ for some integer $k$ since $p>3 \Rightarrow 2\nmid p$ . $p^2(p^2+1)-2=(4k\pm1)^2((4k\pm1)^2+1)-2$ $=(8(2k^2\pm k)+1)(8(2k^2\pm k)+2)-2$ $=8(8(2k^2\pm k)^2+3(2k^2\pm k))$ which is divisible by 8 since $k$ is an integer.

Therefore for all primes $p>3$ , $72 \mid p^4+p^2-2$ so the only primes that satisfy the conditions are 2 and 3 so the product required is $2 \times 3 = 6$

Notice that the expression can be factored into $(p-1)(p+1)(p^2+2).$ This is much easier to examine than our original expression.

Let's check $p=2.$ Obviously, $2^4+2^2-2=18$ is not a multiple of $72.$

For any other prime $p,$ both $p-1$ and $p+1$ must be even, and one must be a multiple of four, so we have a factor of 8 here.

Now let's check $p=3$ . We see that $3^4+3^2-2=88$ is also not a multiple of $72.$

Since any other prime $p$ is not a multiple of three, either $p-1$ or $p+1$ is a multiple of three, since they are three consecutive numbers.

Thus, we've reduced the problem to finding all primes $p$ such that $p^2+2$ is not a multiple of 3. If we examine squares modulo 3, we see that the only residues are 0 and 1. We already checked $p=3$ . For any other prime $p,$ the residue is 1, upon adding 2 we get 0 modulo 3.

Since we've shown that for all primes $p$ (other than 2 and 3) our expression is divisible by 72, our answer is then $2\times3=\boxed{6}.$

p=2,3 is OK if p>=5, 8|p^2-1, 3|(p-1)(p+1), 3|p^2+2, so 72|(p^2+2)(p^2-1)

Note that $p^4+p^2-2=(p^2+2)(p+1)(p-1)$ .

We first consider the special case where $p=2$ because $p$ is even. Plugging it in, we get that $p^4+p^2-2=18$ , which is not a multiple of $72$ .

Now, the rest of the cases must satisfy $p=2n+1$ with $n$ a positive integer.

Plugging in to $(p+1)(p-1)$ , we have $4n^2+4n=4n(n+1)$ . Note that since $n(n+1)$ is always even, then $8|(p+1)(p-1)$ .

We now look at $p^2+2$ . Note that since $p^2\equiv 1\pmod{3}$ because $p$ is odd, then $3|p^2+2$ .

Now, we need to find the $p$ such that $72\not |(p^2+2)(p+1)(p-1)$ . Since we already proved that $24|(p^2+2)(p+1)(p-1)$ , we need to prove that there are no more multiples of $3$ in it. Going back to $4x(x+1)$ , we see that the only way that $3\not |4x(x+1)$ is when $x\equiv 1\pmod{3}\implies x=3a+1$ . Solving for $p$ gives $p=2(3a+1)+1=6a+3$ , so $3|p$ . But the only prime that satisfies this is $3$ .

Therefore, the only two primes that work are $p=2,3$ and the answer is $2\times 3=\boxed{6}$ .

By simply subbing $p=2$ into the expression, it equates to $18$ , which is not a multiple of $72$ , making $2$ a solution.

Looking at $p > 2$ ,

$p^4+p^2-2=(p^2-1)(p^2+2)$

When $p>2$ , it is odd. Hence, we can generalise $p=2k+1$ , where $k$ is any positive integer that makes $2k+1$ a prime.

$(p^2-1)(p^2+2)$

$=([2k+1]^2-1)([2k+1]^2+2)$

$=([4k^2+4k+1]-1)([4k^2+4k+1]+2)$

$=(4k^2+4k)(4k^2+4k+3)$

$=4k(k+1)(4k^2+4k+3)$

Note that $72=2^3 \times 3^2$ . Hence, the expression must not have the factors $2^3 \times 3^2$ in it.

From the equation above, observe that it already has a factor of $4$ . $k(k+1)$ will always have a even number, in other words, a factor of $2$ . Hence, for any value of $k$ , the expression will be a multiple of $8$ . Therefore, the only thing left to find is when $k(k+1)(4k^2+4k+3)$ is not a multiple of $9$ .

$k(k+1)(4k^2+4k+3)$

$=k(k+1)(4k[k+1]+3)$

Obviously, if any of the terms are multiples of $9$ , the entire term will be a multiple of $9$ . Checking for cases when two terms contain factors of $3$ each, Observe that if $k(k+1) \equiv 0 \pmod{3}$ , then $4k(k+1)+3 \equiv 0 \pmod{3}$ as well, making the expression a multiple of $9$ . Thus, the only way in which the expression is not a multiple of $9$ is when $k(k+1)$ is not a multiple of $3$ , occuring when $k \equiv 1 \pmod{3}$ .

$k \equiv 1 \pmod{3}$

$2k+1 \equiv 0 \pmod{3}$

$p \equiv 0 \pmod{3}$

Thus, $p$ must be a multiple of $3$ . The only prime number that is a multiple of $3$ is $3$ itself, making $3$ the only other solution.

Therefore, $2 \times 3= \boxed {6}$

We will use the fact that for every prime $p \ge 5, p^{2} \equiv 1(mod 24)$ Proof: A very popular fact says that for every prime $p \ge 5, p \equiv 1,5(mod6).$ $p^{2}-1=(p+1)(p-1).$ Suppose $p=6k+1$ $\Rightarrow (p+1)(p-1) = 12k(3k+1) \equiv 0 (mod 12).$ Suppose $k$ is even then the above expression is divisible by 24, which is to be proved. Suppose $k$ is odd then $3k+1 \equiv 0(mod2)$ , hence again the above expression is divisible by 24.By the same logic we can prove for $p=6k-1.$ Hence proved.

Observe the $p^{4}+p^{2}-2 = (p^{2} -1)(p^{2} +2).$ Let us assume that $p \ge5$ then $p^{4}+p^{2}-2 \equiv 0 (mod24)$ , Also $p^{2} \equiv 1(mod3)$ , so $(p^{2} -1)(p^{2} +2) \equiv 0 (mod 9)$ Hence the expression is divisible by 72. So, $p <5$ , $\Rightarrow p=2,3$ , and both of then satisfy our conditions, hence the answer.

Factor it as (p-1)(p+1)(p^2+2). Assuming p is not 2 or 3, (p-1)(p+1) must be divisible by 8 and 3. Since p is 1 mod 3, p^2+2 gives us a factor of 3. However, 2 and 3 do not work so the product is 6

Define the function $f(p) = p^4 + p^2 - 2.$ Note that $f(2) = 18$ and $f(3) = 88,$ so the primes $2$ and $3$ do work. We now show that all other primes fail; that is, $72 \mid f(p)$ for all other primes $p.$ It is well-known that for all other primes $p \not\in \{2, 3\},$ we must have either $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6.$ We take cases:

1. $p \equiv 1 \pmod 6$

Notice that $\begin{aligned} f(p) &= p^4+p^2-2 \\ &= (p^2+2)(p^2-1) \\ &= (p^2+2)(p+1)(p-1). \end{aligned}$ Since $p$ is odd, $p-1$ and $p+1$ are consecutive even integers. This means that one must be divisible by $4$ and the other must be divisible by $2,$ therefore $f(p)$ contains at least three factors of $2.$ We also see that $p^2 + 2 \equiv 1^2 + 2 = 3 \pmod 6$ and $p - 1 \equiv 1 -1 = 0\pmod 6,$ which means that both $p^2 + 2$ and $p-1$ are multiples of $3.$ Therefore, $f(p)$ contains at least two factors of $3.$ We can then conclude that $f(p)$ is a multiple of $2^3 \cdot 3^2 = 72.$

1. $p \equiv 5 \pmod 6$

We again apply our factorization $f(p) = (p^2+2)(p+1)(p-1).$ Similarly, since $p+1$ and $p-1$ are consecutive even integers, one is divisible by $4$ and the other is divisible by $2,$ so $8 \mid f(p).$ In addition, $p^2 + 2 \equiv 5^2 + 2 \equiv 3 \pmod 6$ and $p+1 \equiv 5 + 1\equiv 0 \pmod 6,$ so $9 \mid f(p)$ as well. Again, we conclude that $72 \mid f(p).$

Thus, the only primes such that $72 \not\mid \, f(p)$ are $p = 2, 3,$ which multiply to give our answer, $\boxed{6}.$

p^4+p^2-2=(p^2-1)(p^2+2) .... taking (p^2-1) 1st , we know for p>2 p is always odd prime which can be written as 2k+1 .... thus (p^2-1) becomes 4k(k+1) which is always a multiple of 8 as k(k+1) is always a multiple of 2 .. now taking (p^2+2) let p is a prime number other then 3 then we can represent p= 3k+1 or 3k+2(for some k and not all k) that is remainder with respect to 3 is always 1 or 2(-1).. making (p^2)mod3=1 and 1 +2=3 thus (p^2+2) is always a multiple of 3. now since (p^2+2)-(p^2-1) is always =3 thus (p^2-1) is also a multiple of 3. that is for p>3 (p^2-1)=24k and (p^2+2)=3k thus (p^2+2)(p^2-1)=72k that is a multiple of 72. only remaining prime numbers to be checked are 3 or 2 for which it is not a multiple of 72 and hence answer is=3*2=6

This can be factored as $p^4 + p^2 - 2 = (p - 1)(p + 1)(p^2 + 2)$ . These three factors must share at least $3$ factors of $2$ and $2$ factors of $3$ .

For odd $p$ , $p - 1$ and $p + 1$ have at least three factors of two between them they're both even, and since they're consecutive then one must have a factor of $4$ as well. For $p = 2$ , there is an insufficient amount of twos, so we can write that out.

Finally, $p$ can either be congruent to $0, 1, 2 \pmod 3$ . For $p \equiv 0$ , there are no factors of three, thus we can add $p = 3$ to the list. For all other $p$ , there is at least one factor of three in $(p+1)(p-1) = p^2 - 1$ , thus there must also be a factor of three in $p^2 - 1 + 3 = p^2 + 2$ . Thus, for all other primes, there are at least two factors of three and three factors of two, making our product $2 \times 3 = 6$ .

Let $f(p):=p^4+p^2-2$ We will show that for any $p\ge5$ , $f(p)\equiv0\pmod{72}$ . We start factoring the polynomial: $f(p)=(p^2-1)(p^2+2)$ Then we split the congruence in two parts, by CRT.

• $f(p)\equiv0\pmod{8}$ It is known and easily checked by hand that for every odd $n$ we have that $n^2-1\equiv0\pmod8$ ; now, we know that $p\ge5$ , so $p$ is odd. Then $f(p)\equiv(p^2-1)(p^2+2)\equiv0\cdot(p^2+2)\equiv0\pmod8$ .

• $f(p)\equiv0\pmod{9}$ Since $p\ge5$ , we have that $p\equiv\pm1\pmod3$ , so $p^2\equiv1\pmod3$ . Then $p^2-1\equiv1-1\equiv0\pmod3$ and $p^2+2\equiv1+2\equiv0\pmod3$ But then $f(p)$ is the product of two multiples of $3$ , so it's a multiple of $9$ .

Now, the only possibilities for $p$ are $2,3$ ; we have that $f(2)=18$ and $f(3)=88$ ; they aren't multiple of $72$ , so our product is $2\cdot3=\boxed6$ .

$72=8*9$

Firstly since every odd perfect square is $1 \mod 8$ . $p^4+p^2-2=1+1-2=0 \mod 8$ unless p=2.

Now,write the expression as $p^2(p^2+1)-2$ Then, $p^2$ can have values $1,4,7$ in mod $9$ only unless p=3. Easy to check that they all implies the expression is divisible by $9$ unless p=3.

So, unless $p$ is $2$ or $3$ the expression is divisible by 72.

Hence,the answer is $2*3=\boxed{6}$

For $p=2$ and $p=3$ , $p^4+p^2-2$ is clearly not a multiple of 72. Now, we need to consider the rest of the primes.

Let $p=2x+1$ , where $x$ is an integer. Substituting $2x+1$ for $p$ , we get $p^4+p^2-2=4x(4x^3+8x^2+7x+3)$ Whether $x$ is even or odd, the expression will be divisble by 8. To check for divisibility by 9, we check all residues modulo 3. If $x \equiv 0 \pmod{3}$ then the expression is divisible by 9. If $x \equiv 2 \pmod{3}$ then it is also divisble by 9. If $x \equiv 1 \pmod{3}$ the expression is not divisible by 9. However, if $x \equiv 1 \pmod{3}$ then $x \equiv 0 \pmod{3}$ which is impossible for primes greater than 3. Therefore, all primes greater than 3 are divisible by 72 and our answer is $2 \times 3=6$ .