The question should ask I suspect that there are infinitely many exceptions however this would imply that that there are infinitely many Sophie Germain primes (2p+1 also prime). This is not known to be true although there is every reason to expect that there are about $\frac{n}{\log^2n}$ such less than $n$.

$p=(k+1)(ab)+k(a+b)$ or equivalently is true exactly if $p\equiv (ab)\bmod ((a+b)+ab)$ with (k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number$\mathbf{ab<p}$Otherwise you are allowing (k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $k=0,a=1,b=p$.-1 \mod k+1$The largest that$k$could be is$\frac{p-2}{5}$in case$(a,b)=(1,1)$. Of the 1229 course then$2p+1$is not prime (if$p \ne 2$) It may be that there are only finitely many primes with$(k+1)p+k^2$prime for all$k$up to 10000, 190 are Sophie Germain primes$p,2p+1$both p/5$ however we don't need them all prime. , just to not have any factor congruent to $k \mod k+1.$

The first few Sophie Germain primes are

If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.

later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$

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If $2p+1$ is composite there is a solution with $k=1$ and if it is prime then there is none with $k=1$. In general there is a solution with $k=K$ if and only if $(K+1)p+K^2$ factors as the product of two factors both congruent to -1 $\mod K$.

$p=(k+1)(ab)+k(a+b)$ or equivalently $p\equiv (ab)\bmod ((a+b)+ab)$ with $\mathbf{ab<p}$

Otherwise you are allowing $k=0,a=1,b=p$.

Of the 1229 primes up to 10000, 190 are Sophie Germain primes $p,2p+1$ both prime. The first few are

$$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173, 179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above except 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$

In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. All of the exceptions are Sophie Germain primes. The largest Sophie Germain primes under 1000 are

$$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with the exceptions of 9371 and 9539.

If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.

later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$

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$p=(k+1)(ab)+k(a+b)$ or equivalently $p\equiv (ab)\bmod ((a+b)+ab)$ with $\mathbf{ab<p}$

Otherwise you are allowing $k=0,a=1,b=p$.

Of the 1229 primes up to 10000, 190 are Sophie Germain primes $p,2p+1$ both prime. The first few are

$$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173, 179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above except 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$

In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. All of the exceptions are Sophie Germain primes. The largest Sophie Germain primes under 1000 are

$$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with the exceptions of 9371 and 9539.

If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.

later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$

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