$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
what I get is
$$ a = - p r + qs $$ $$ b = pq + 3 rs + 2 p r + q s $$ $$ c = 3 rs + p r + q s $$ $$ d = pq + p r + q s $$
after which $$ a+b+c+d = (p+2s)(2q+3r). $$
I picked this one as having the fewest available minus signs.
The remaining part is why we never get $p+2s = \pm 1$ or $2q+3r = \pm 1$ and the other factor prime. The rules are: $a,b,c,d$ positive, $\gcd(a,b,c,d) = 1,$ $(c,d) \neq (a,b)$ and $(c,d) \neq (b,a)$ and probably others. Since I am demanding $a^2 + ab + b^2$ have at least two essentially different expressions with positive variables, it cannot be a prime or three times a prime. I see: it can be a prime cubed but not a prime squared, properties of Gauss composition following from class number one.
SLOGGING through: we want to rule out any chance of the product $ (p+2s)(2q+3r)$ being prime, using the conditions above in a minimal way. For example, if $p+2s = 1,$ we cannot have $a,b,c,d$ all positive, as then $a+c= (2q+3r)s$ while $b+d = (2q+3r)(1-s).$ As $a+b+c+d \neq 0,$ we know $(2q+3r) \neq 0.$ Either $s \leq 0$ or $(1-s) \leq 0,$ so either $a+c \leq 0$ or $b+d \leq 0,$ not allowed. We have shown that $p+2s \neq 1.$
Three more to go, $p+2s \neq -1,$ $2q+3r \neq 1,$ $2q+3r \neq -1$
a^2 + ab + b^2 a b c d
91 = 7 13 ; 1 9 6 5 sum 21 = 3 7 pqrs 1 2 1 1
91 = 7 13 ; 5 6 9 1 sum 21 = 3 7 pqrs 1 -2 -1 -2
91 = 7 13 ; 6 5 1 9 sum 21 = 3 7 pqrs 1 5 -1 1
91 = 7 13 ; 9 1 5 6 sum 21 = 3 7 pqrs 1 -5 1 -2
133 = 7 19 ; 11 1 4 9 sum 25 = 5^2 pqrs 1 -4 1 -3
133 = 7 19 ; 1 11 9 4 sum 25 = 5^2 pqrs 1 1 1 2
133 = 7 19 ; 4 9 11 1 sum 25 = 5^2 pqrs 1 -1 -1 -3
133 = 7 19 ; 9 4 1 11 sum 25 = 5^2 pqrs 1 4 -1 2
147 = 3 7^2 ; 2 11 7 7 sum 27 = 3^3 pqrs 1 3 1 1
147 = 3 7^2 ; 7 7 11 2 sum 27 = 3^3 pqrs 1 -3 -1 -2
217 = 7 31 ; 3 13 8 9 sum 33 = 3 11 pqrs 1 4 1 1
217 = 7 31 ; 9 8 13 3 sum 33 = 3 11 pqrs 1 -4 -1 -2
247 = 13 19 ; 11 7 3 14 sum 35 = 5 7 pqrs 1 5 -1 2
247 = 13 19 ; 14 3 7 11 sum 35 = 5 7 pqrs 1 -5 1 -3
247 = 13 19 ; 3 14 11 7 sum 35 = 5 7 pqrs 1 2 1 2
247 = 13 19 ; 7 11 14 3 sum 35 = 5 7 pqrs 1 -2 -1 -3
259 = 7 37 ; 13 5 2 15 sum 35 = 5 7 pqrs 1 4 -1 3
259 = 7 37 ; 15 2 5 13 sum 35 = 5 7 pqrs 1 -4 1 -4
259 = 7 37 ; 2 15 13 5 sum 35 = 5 7 pqrs 1 1 1 3
259 = 7 37 ; 5 13 15 2 sum 35 = 5 7 pqrs 1 -1 -1 -4
273 = 3 7 13 ; 1 16 11 8 sum 36 = 2^2 3^2 pqrs 1 3 2 1
273 = 3 7 13 ; 1 16 8 11 sum 36 = 2^2 3^2 pqrs 2 3 1 1
273 = 3 7 13 ; 11 8 16 1 sum 36 = 2^2 3^2 pqrs 2 -3 -1 -3
273 = 3 7 13 ; 8 11 16 1 sum 36 = 2^2 3^2 pqrs 1 -3 -2 -2
301 = 7 43 ; 11 9 15 4 sum 39 = 3 13 pqrs 1 -5 -1 -2
301 = 7 43 ; 4 15 9 11 sum 39 = 3 13 pqrs 1 5 1 1
343 = 7^3 ; 1 18 14 7 sum 40 = 2^3 5 pqrs 2 1 1 3
343 = 7^3 ; 14 7 1 18 sum 40 = 2^3 5 pqrs 2 4 -1 3
343 = 7^3 ; 18 1 7 14 sum 40 = 2^3 5 pqrs 2 -4 1 -5
343 = 7^3 ; 7 14 18 1 sum 40 = 2^3 5 pqrs 2 -1 -1 -5
399 = 3 7 19 ; 10 13 17 5 sum 45 = 3^2 5 pqrs 1 -3 -1 -3
399 = 3 7 19 ; 5 17 13 10 sum 45 = 3^2 5 pqrs 1 3 1 2
403 = 13 31 ; 14 9 19 2 sum 44 = 2^2 11 pqrs 2 -4 -1 -3
403 = 13 31 ; 2 19 9 14 sum 44 = 2^2 11 pqrs 2 4 1 1
427 = 7 61 ; 17 6 3 19 sum 45 = 3^2 5 pqrs 1 4 -1 4
427 = 7 61 ; 19 3 6 17 sum 45 = 3^2 5 pqrs 1 -4 1 -5
427 = 7 61 ; 3 19 17 6 sum 45 = 3^2 5 pqrs 1 1 1 4
427 = 7 61 ; 6 17 19 3 sum 45 = 3^2 5 pqrs 1 -1 -1 -5
469 = 7 67 ; 12 13 20 3 sum 48 = 2^4 3 pqrs 1 -5 -2 -2
469 = 7 67 ; 3 20 13 12 sum 48 = 2^4 3 pqrs 1 5 2 1
481 = 13 37 ; 16 9 5 19 sum 49 = 7^2 pqrs 1 5 -1 3
481 = 13 37 ; 19 5 9 16 sum 49 = 7^2 pqrs 1 -5 1 -4
481 = 13 37 ; 5 19 16 9 sum 49 = 7^2 pqrs 1 2 1 3
481 = 13 37 ; 9 16 19 5 sum 49 = 7^2 pqrs 1 -2 -1 -4
507 = 3 13^2 ; 1 22 13 13 sum 49 = 7^2 pqrs 3 2 1 2
507 = 3 13^2 ; 13 13 1 22 sum 49 = 7^2 pqrs 3 5 -1 2
507 = 3 13^2 ; 13 13 22 1 sum 49 = 7^2 pqrs 3 -2 -1 -5
507 = 3 13^2 ; 22 1 13 13 sum 49 = 7^2 pqrs 3 -5 1 -5
553 = 7 79 ; 11 16 23 1 sum 51 = 3 17 pqrs 1 -4 -3 -2
553 = 7 79 ; 1 23 16 11 sum 51 = 3 17 pqrs 1 4 3 1
559 = 13 43 ; 17 10 22 3 sum 52 = 2^2 13 pqrs 2 -5 -1 -3
559 = 13 43 ; 3 22 10 17 sum 52 = 2^2 13 pqrs 2 5 1 1
589 = 19 31 ; 13 15 20 7 sum 55 = 5 11 pqrs 1 -4 -1 -3
589 = 19 31 ; 7 20 15 13 sum 55 = 5 11 pqrs 1 4 1 2
637 = 7^2 13 ; 12 17 23 4 sum 56 = 2^3 7 pqrs 2 -2 -1 -5
637 = 7^2 13 ; 17 12 4 23 sum 56 = 2^3 7 pqrs 2 5 -1 3
637 = 7^2 13 ; 21 7 4 23 sum 55 = 5 11 pqrs 1 4 -1 5
637 = 7^2 13 ; 23 4 12 17 sum 56 = 2^3 7 pqrs 2 -5 1 -5
637 = 7^2 13 ; 4 23 17 12 sum 56 = 2^3 7 pqrs 2 2 1 3
637 = 7^2 13 ; 4 23 21 7 sum 55 = 5 11 pqrs 1 1 1 5
651 = 3 7 31 ; 1 25 10 19 sum 55 = 5 11 pqrs 3 4 1 1
651 = 3 7 31 ; 1 25 19 10 sum 55 = 5 11 pqrs 3 1 1 4
651 = 3 7 31 ; 19 10 1 25 sum 55 = 5 11 pqrs 3 4 -1 4
651 = 3 7 31 ; 19 10 25 1 sum 55 = 5 11 pqrs 3 -4 -1 -4
679 = 7 97 ; 13 17 25 2 sum 57 = 3 19 pqrs 1 -5 -3 -2
679 = 7 97 ; 2 25 17 13 sum 57 = 3 19 pqrs 1 5 3 1
703 = 19 37 ; 1 26 23 6 sum 56 = 2^3 7 pqrs 1 1 2 3
703 = 19 37 ; 6 23 26 1 sum 56 = 2^3 7 pqrs 1 -1 -2 -4
741 = 3 13 19 ; 11 20 25 4 sum 60 = 2^2 3 5 pqrs 1 -3 -2 -3
741 = 3 13 19 ; 4 25 20 11 sum 60 = 2^2 3 5 pqrs 1 3 2 2
763 = 7 109 ; 22 9 3 26 sum 60 = 2^2 3 5 pqrs 2 4 -1 5
763 = 7 109 ; 3 26 22 9 sum 60 = 2^2 3 5 pqrs 2 1 1 5
777 = 3 7 37 ; 13 19 23 8 sum 63 = 3^2 7 pqrs 1 -3 -1 -4
777 = 3 7 37 ; 8 23 19 13 sum 63 = 3^2 7 pqrs 1 3 1 3
793 = 13 61 ; 11 21 24 7 sum 63 = 3^2 7 pqrs 1 -2 -1 -5
793 = 13 61 ; 21 11 7 24 sum 63 = 3^2 7 pqrs 1 5 -1 4
793 = 13 61 ; 24 7 11 21 sum 63 = 3^2 7 pqrs 1 -5 1 -5
793 = 13 61 ; 7 24 21 11 sum 63 = 3^2 7 pqrs 1 2 1 4
817 = 19 43 ; 16 17 23 9 sum 65 = 5 13 pqrs 1 -5 -1 -3
817 = 19 43 ; 9 23 17 16 sum 65 = 5 13 pqrs 1 5 1 2
871 = 13 67 ; 1 29 15 19 sum 64 = 2^6 pqrs 2 5 2 1
871 = 13 67 ; 19 15 29 1 sum 64 = 2^6 pqrs 2 -5 -2 -3
903 = 3 7 43 ; 2 29 11 23 sum 65 = 5 13 pqrs 3 5 1 1
903 = 3 7 43 ; 2 29 23 11 sum 65 = 5 13 pqrs 3 1 1 5
903 = 3 7 43 ; 23 11 2 29 sum 65 = 5 13 pqrs 3 4 -1 5
903 = 3 7 43 ; 23 11 29 2 sum 65 = 5 13 pqrs 3 -5 -1 -4
931 = 7^2 19 ; 1 30 21 14 sum 66 = 2 3 11 pqrs 1 5 4 1
931 = 7^2 19 ; 1 30 25 9 sum 65 = 5 13 pqrs 1 2 3 2
931 = 7^2 19 ; 14 21 30 1 sum 66 = 2 3 11 pqrs 1 -5 -4 -2
931 = 7^2 19 ; 9 25 30 1 sum 65 = 5 13 pqrs 1 -2 -3 -3
1027 = 13 79 ; 19 18 2 31 sum 70 = 2 5 7 pqrs 4 5 -1 3
1027 = 13 79 ; 2 31 19 18 sum 70 = 2 5 7 pqrs 4 2 1 3
1029 = 3 7^3 ; 17 20 28 7 sum 72 = 2^3 3^2 pqrs 2 -3 -1 -5
well, why not. Using C++ with GMP, here is the active part of the program:
mpz_class bound = 5;
for(mpz_class p = 0; p <= bound; ++p){
for(mpz_class q = -bound; q <= bound; ++q){
for(mpz_class r = -bound; r <= bound; ++r){
for(mpz_class s = -bound; s <= bound; ++s){
mpz_class a = q * s - p * r;
mpz_class b = p*q + 3 * r*s + 2 * p * r + q * s ;
mpz_class c = 3 * r*s + p * r + q * s ;
mpz_class d = p*q + p * r + q * s ;
if ( a > 0 && b> 0 && c>0 && d > 0 && !( (a == c && b == d) || (a == d && b == c) ) && mp_coprime4(a,b,c,d) )
{
cout << setw(8) << a *a+a*b+b*b<< " = " << mp_Factored(a*a+a*b+b*b) << " ; " << setw(3) << a << setw(3) << b << setw(3) << c << setw(3) << d << " sum " << setw(4) << a +b+c+d << " = " << mp_Factored(a+b+c+d)<< " pqrs " << setw(3) << p << setw(3) << q << setw(3) << r << setw(3) << s << endl;
if (mp_PrimeQ(a+b+c+d) ) cout << " WOE " << endl;
}
}}}}
We work in the principal ideal domain $\mathbb Z [\omega]$, where $\omega = \frac{-1 + \sqrt3 i}{2}$ (which we view as an element of $\mathbb C$).
Without loss of generality, assume that $a \geq b$ and $c \geq d$, and that we don't have equality at the same time.
Write the equation as $(a + b \omega)(a + b\overline\omega) = (c + d \omega)(c + d\overline\omega)$. Let $u$ be the greatest common divisor of $a + b\omega$ and $c + d\omega$. It is well defined up to multiplication by a sixth root of unity, i.e. $\pm 1, \pm \omega, \pm \omega^2$.
Write $a + b\omega = u v$ and $c + d\omega = u v'$, with $v, v'$ coprime. We then have $v\overline v = v' \overline {v'}$, which implies (since $v, v'$ coprime) that $v \mid \overline{v'}$ and $v'\mid \overline v$. But taking complex conjugation gives $\overline{v'}\mid v$, hence $v' = \varepsilon v$ for some sixth root of unity $\varepsilon$.
We therefore have $a + b\omega = uv$, $c + d\omega = u\overline v \varepsilon$. By replacing $u, v, \varepsilon$ with $u\delta^{-1}, v\delta, \varepsilon\delta^2$ for some sixth root of unity $\delta$, we may assume without loss of generality that $\varepsilon = \pm 1$.
Case 1: $\varepsilon = 1$.
We have $a + b\omega = uv$. Taking complex conjugation gives $a + b\overline \omega = \overline u\overline v$. Solving this linear equation, we get $a + b = -\omega u v - \overline \omega \overline u \overline v$.
Similarly, we get $c + d = -\omega u \overline v - \overline \omega \overline u v$.
Hence in the end $a + b + c + d = -(\omega u + \overline {\omega u})(v + \overline v)$.
Since both $(\omega u + \overline {\omega u})$ and $v + \overline v$ are integers, if the product is prime, then one of them must be $\pm 1$.
Now observe the fact that, for any element $x \in \mathbb Z[\omega]$, $x + \overline x = \pm 1$ will imply that $\arg x$ lies in either $[60^\circ, 120^\circ]$ or $[-120^\circ, -60^\circ]$.
It follows that $v + \overline v$ cannot be $\pm 1$. This is because both $a + b\omega$ and $c + d\omega$ lives in the region $\{z \in \mathbb C: 0 < \arg z \leq 60^\circ\}$, hence the number $v/\overline v$, being their quotient, must have $\arg$ in the range $(-60^\circ, 60^\circ)$.
But $\omega u + \overline{\omega u}$ cannot be $\pm 1$ either. The reasoning is similar: since $a + b\omega$ and $c + d\omega$ cannot both attain $60^\circ$ of $\arg$, the number $u^2 v \overline v$, being their product, must have $\arg$ in the range$(0^\circ, 120^\circ)$. This means $u$ must have $\arg$ in the range $(0^\circ, 60^\circ)$ or $(-180^\circ, -120^\circ)$, hence $\omega u$ has $\arg$ in the range $(120^\circ, 180^\circ)$ or $(-60^\circ, 0^\circ)$.
This completes the proof that $a + b + c + d$ cannot be a prime number in the case $\varepsilon = 1$.
Case 2: $\varepsilon = -1$.
Same as above, we work out the formula $a + b + c + d = -(v - \overline v)(\omega u - \overline{\omega u})$.
This case is now significantly easier, since the product must be a multiple of $3$. So if it is a prime, then we have $a + b + c + d = 3$ and it's already impossible.
If $ \{ a, b \} = \{c, d \}$ then $a+b+c+d = 2(a+b)$ is not a prime.
Henceforth, we have distinct pairs. If $ a = c$, then $b \neq d$ are roots to $x^2 + ax + a^2 = 0$, so $b+d = -a$ contradicting the requirement that all of the terms are positive.
WLOG, $ a > c \geq d > b$.
Observe that $$ -(a-b+c-d)(a-b-c+d) = -\left((a-b)^2 - (c-d)^2\right) = 3(ab-cd) = 3\left( (a+b)^2 - (c+d)^2\right) = 3 (a+b-c-d)(a+b+c+d).$$
Since $ a+b+c+d > a-b+c-d \geq a-b-c+d > 0$, we can rewrite the above as a product of positive integers,
$$ (a-b+c-d)(a-b-c+d)= 3 (c+d-a-b)(a+b+c+d)$$
If $a+b+c+d$ is prime, then the LHS can never be a multiple of $a+b+c+d$ since both terms are smaller, hence we have a contradiction. So $ a+b+c+d$ is composite.
Notes:
- This problem/solution is very reminiscent of 2001/6 IMO. There we had $a^2+ab+b^2 = c^2 -cd+d^2$.
- I do not know a direct way to show that $cd > ab$. Any thoughts? This seems obvious.
- This being a contest problem suggests that there is a non "abstract algebra approach", even though it is very tempting. E.g. The equation in my solution can also be derived from WhatsUp's characterization (which is a stronger requirement).