How can I solve the following diophantine equation ?
$$16c^2=a^2+15b^2$$ where $a,b,c\in\mathbb Z$ .
We are not interested in the trivial cases $a=b=c$ or $b=0$ .
$\rm {Attempts :}$ ( Left for the Original Poster )
2026-04-05 23:19:02.1775431142
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Solving the equation $16c^2=a^2+15b^2$ over integers
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Posting what I meant: six Pythagorean Triple type parametrizations give all primitive integer solutions to $a^2 + 15 b^2 = 16 c^2$ For example, the one that gives $(1,1,1) $ is capital letter B: $$ a = u^2 - 30uv - 15 v^2 \; ; \; \; b = u^2 +2uv - 15 v^2 \; ; \; \; c = u^2 + 15 v^2 \; ; \; \; $$
Page 47 from Mordell, Diophantine Equations:
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
a b c u v
1 1 1 B 1 0
1 1 1 B -1 0
1 1 1 RAW
1 63 61 B 1 -2
1 63 61 B -1 2
1 63 61 RAW
4 0 1 D 1 0
4 0 1 D -1 0
4 0 1 RAW
4 32 31 D 0 1
4 32 31 D 0 -1
4 32 31 D 8 -1
4 32 31 D -8 1
4 32 31 RAW
7 1 2 E 1 0
7 1 2 E -1 0
7 1 2 RAW
7 33 32 E 1 1
7 33 32 E -1 -1
7 33 32 RAW
7 95 92 E 11 -3
7 95 92 E -11 3
7 95 92 RAW
11 221 214 F 8 7
11 221 214 F -8 -7
11 221 214 RAW
11 3 4 F 1 0
11 3 4 F -1 0
11 3 4 RAW
11 35 34 F 2 -1
11 35 34 F -2 1
11 35 34 RAW
17 39 38 E 7 -2
17 39 38 E -7 2
17 39 38 RAW
17 7 8 E 3 -1
17 7 8 E -3 1
17 7 8 RAW
28 16 17 C 2 1
28 16 17 C 2 -1
28 16 17 C -2 1
28 16 17 C -2 -1
28 16 17 RAW
28 48 47 C 3 2
28 48 47 C 3 -2
28 48 47 C -3 2
28 48 47 C -3 -2
28 48 47 RAW
43 21 23 A 1 2
43 21 23 A -1 -2
43 21 23 RAW
43 85 83 A 1 4
43 85 83 A -1 -4
43 85 83 RAW
44 112 109 D 1 -2
44 112 109 D -1 2
44 112 109 D 15 -2
44 112 109 D -15 2
44 112 109 RAW
44 16 19 D 2 -1
44 16 19 D -2 1
44 16 19 D 6 -1
44 16 19 D -6 1
44 16 19 RAW
49 143 139 B 2 -3
49 143 139 B -2 3
49 143 139 RAW
49 15 19 B 2 -1
49 15 19 B -2 1
49 15 19 RAW
53 11 17 A 2 -1
53 11 17 A -2 1
53 11 17 RAW
53 203 197 A 8 -1
53 203 197 A -8 1
53 203 197 RAW
59 45 46 F 4 3
59 45 46 F -4 -3
59 45 46 RAW
59 77 76 F 5 4
59 77 76 F -5 -4
59 77 76 RAW
61 5 16 F 1 2
61 5 16 F -1 -2
61 5 16 RAW
67 3 17 A 2 1
67 3 17 A -2 -1
67 3 17 RAW
68 16 23 C 1 2
68 16 23 C 1 -2
68 16 23 C -1 2
68 16 23 C -1 -2
68 16 23 RAW
68 240 233 C 6 5
68 240 233 C 6 -5
68 240 233 C -6 5
68 240 233 C -6 -5
68 240 233 RAW
71 7 19 B 2 1
71 7 19 B -2 -1
71 7 19 RAW
77 115 113 A 6 -1
77 115 113 A -6 1
77 115 113 RAW
77 13 23 A 1 -2
77 13 23 A -1 2
77 13 23 RAW
77 51 53 A 4 -1
77 51 53 A -4 1
77 51 53 RAW
103 161 158 E 3 2
103 161 158 E -3 -2
103 161 158 RAW
103 65 68 E 3 1
103 65 68 E -3 -1
103 65 68 RAW
109 21 34 F 2 3
109 21 34 F -2 -3
109 21 34 RAW
113 217 212 E 1 3
113 217 212 E -1 -3
113 217 212 RAW
113 57 62 E 1 -2
113 57 62 E -1 2
113 57 62 RAW
119 247 241 B 1 -4
119 247 241 B -1 4
119 247 241 RAW
119 55 61 B 1 2
119 55 61 B -1 -2
119 55 61 RAW
119 9 31 B 4 1
119 9 31 B -4 -1
119 9 31 RAW
121 7 31 B 4 -1
121 7 31 B -4 1
121 7 31 RAW
127 55 62 E 9 -2
127 55 62 E -9 2
127 55 62 RAW
131 187 184 F 5 -2
131 187 184 F -5 2
131 187 184 RAW
131 91 94 F 4 -1
131 91 94 F -4 1
131 91 94 RAW
137 17 38 E 3 -2
137 17 38 E -3 2
137 17 38 RAW
163 35 53 A 4 1
163 35 53 A -4 -1
163 35 53 RAW
172 32 53 C 4 1
172 32 53 C 4 -1
172 32 53 C -4 1
172 32 53 C -4 -1
172 32 53 RAW
173 19 47 A 3 -2
173 19 47 A -3 2
173 19 47 RAW
179 11 46 F 4 1
179 11 46 F -4 -1
179 11 46 RAW
181 99 106 F 2 -3
181 99 106 F -2 3
181 99 106 RAW
187 133 137 A 2 5
187 133 137 A -2 -5
187 133 137 RAW
187 5 47 A 3 2
187 5 47 A -3 -2
187 5 47 RAW
191 65 79 B 8 1
191 65 79 B -8 -1
191 65 79 RAW
196 64 79 D 12 -1
196 64 79 D -12 1
196 64 79 D 4 1
196 64 79 D -4 -1
196 64 79 RAW
197 69 83 A 1 -4
197 69 83 A -1 4
197 69 83 RAW
212 96 107 C 3 4
212 96 107 C 3 -4
212 96 107 C -3 4
212 96 107 C -3 -4
212 96 107 RAW
223 119 128 E 13 -3
223 119 128 E -13 3
223 119 128 RAW
229 77 94 F 4 5
229 77 94 F -4 -5
229 77 94 RAW
233 145 152 E 1 -3
233 145 152 E -1 3
233 145 152 RAW
236 16 61 D 7 -2
236 16 61 D -7 2
236 16 61 D 9 -2
236 16 61 D -9 2
236 16 61 RAW
239 209 211 B 14 1
239 209 211 B -14 -1
239 209 211 RAW
241 143 151 B 4 -3
241 143 151 B -4 3
241 143 151 RAW
244 176 181 D 19 -2
244 176 181 D -19 2
244 176 181 D 3 2
244 176 181 D -3 -2
244 176 181 RAW
a a c u v
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
A:
a = 3*u*u + 30*u*v -5*v*v;
b = 3*u*u -2*u*v -5*v*v;
c = 3*u*u -0*u*v +5*v*v;
B:
a = 1*u*u - 30*u*v -15*v*v;
b = 1*u*u +2*u*v -15*v*v;
c = 1*u*u -0*u*v +15*v*v;
C:
a = 12*u*u + 0*u*v -20*v*v;
b = 0*u*u + 8*u*v + 0*v*v;
c = 3*u*u -0*u*v +5*v*v;
D:
a = 4*u*u + 32*u*v +4*v*v;
b = 0*u*u + 8*u*v + 32*v*v;
c = 1*u*u +8*u*v +31*v*v;
E:
a = 7*u*u + 20*u*v -20*v*v;
b = 1*u*u + 12*u*v + 20*v*v;
c = 2*u*u +10*u*v +20*v*v;
F:
a = 11*u*u + 6*u*v -21*v*v;
b = 3*u*u - 10*u*v + 3*v*v;
c = 4*u*u -6*u*v +6*v*v;
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
This answer can be deleted by voting if the question asked does not improve in the future .
Let $\thinspace b≠0\thinspace$, then you have :
$$ \begin{align}16c^2&=a^2+15b^2\\ 16\left(\frac cb\right)^2&=\left(\frac ab \right)^2+15\end{align} $$
Actually, it is enough to solve the equation for $a,b,c\in\mathbb Z_{≥0}$ . But, we will go to the solution without this restriction .
Let $\thinspace m=\dfrac cb\thinspace $ and $\thinspace n=\dfrac ab\thinspace $, then we can write the following :
$$ \begin{align}&(4m-n)(4m+n)=15\\\\ \implies &\begin{cases} 4m-n=k,\thinspace k\in\mathbb Q_{≠\thinspace 0}\\ 4m+n=\dfrac {15}{k}\end{cases}\\\\ \implies &\begin{cases}m=\dfrac {k^2+15}{8k}\\ n=\dfrac{15-k^2}{2k}\end{cases} \end{align} $$
Let $\thinspace k=\dfrac pq$, where $(p,q)\in\mathbb Z^{2}$ with $q≠0$, then we get :
$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{m=\frac {p^2+15 q^2}{8pq}, \thinspace\thinspace n=\frac {4(15 q^2-p^2)}{8pq}}\end{align} $$
Finally, recall that you can always write $m$ and $n$ in the common denominator. One possible general explicit solution family can be constructed as follows :
$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{\begin{align}a&=4r(15q^2-p^2)\\ b&=8pqr\\ c&=r(p^2+15q^2) \end{align}}\end{align} $$
where $(p,q,r)\in\mathbb Z^{3}$ with $p,q,r≠0\thinspace .$
For example, let $k=\dfrac 35$ we have :
$$ \begin{align}m&=\frac cb=\frac {16}{5}\\ n&=\frac ab=\frac {61}{5}\end{align} $$
which leads to :
$$a=61k,\thinspace b=5k\thinspace ,c=16k$$
where $k\in\mathbb Z\thinspace .$