35 	unsigned m;  in mkrand()  local
42 	m = esize & 15;  in mkrand()
43 	if (m == 0) {  in mkrand()
46 		x[len] &= 0x7FFF >> (15 - m);  in mkrand()
276 		 * Maximum values of m* at this point:  in mkprime()
351 		} else if (esize < 480) {  in mkprime()
370  * Let p be a prime (p > 2^33, p = 3 mod 4). Let m = (p-1)/2, provided
380 invert_pubexp(uint16_t *d, const uint16_t *m, uint32_t e, uint16_t *t)  in invert_pubexp()  argument
386 	t += 1 + ((m[0] + 15) >> 4);  in invert_pubexp()
389 	 * Compute d = 1/e mod m. Since p = 3 mod 4, m is odd.  in invert_pubexp()
391 	br_i15_zero(d, m[0]);  in invert_pubexp()
393 	br_i15_zero(f, m[0]);  in invert_pubexp()
397 	r = br_i15_moddiv(d, f, m, br_i15_ninv15(m[1]), t);  in invert_pubexp()
400 	 * We really want d = 1/e mod p-1, with p = 2m. By the CRT,  in invert_pubexp()
401 	 * the result is either the d we got, or d + m.  in invert_pubexp()
403 	 * Let's write e*d = 1 + k*m, for some integer k. Integers e  in invert_pubexp()
404 	 * and m are odd. If d is odd, then e*d is odd, which implies  in invert_pubexp()
405 	 * that k must be even; in that case, e*d = 1 + (k/2)*2m, and  in invert_pubexp()
407 	 * is odd, and we must add m to d in order to get the correct  in invert_pubexp()
410 	br_i15_add(d, m, (uint32_t)(1 - (d[1] & 1)));  in invert_pubexp()