Special Triangle

Question

A triangle with integer sides a, b and 13 is given so that a is also a prime. The angle opposite to c=13 is 120° (m(C)=120°).

Find a and b.

The answer is a=7 and b=8. Howewer, how can we solve it using modular arithmetical analysis without trying immediate integer values?

Answer 1

Apply the law of cosines.

You must have that

$$(13)^2 = a^2 + b^2 - 2ab[\cos(120^\circ)] \implies $$

$$(13)^2 = a^2 + b^2 + ab = (a + b)^2 - ab. \tag1 $$

This implies that $~(a + b + 13) \times (a + b - 13) = ab.$

Since $~a~$ is prime, $~a~$ must divide either $~(a + b + 13),~$ or $~(a + b - 13).~$ Further, it is immediate from (1) above, that both $~a~$ and $~b~$ are less than $~13.$

Therefore, $~a > a + b - 13 > 0.$

Therefore, the prime $~a~$ must divide $~a + b + 13.$

Therefore, the prime $~a~$ must divide $~b + 13.~$

Further, $~a + b - 13 > 0 \implies (a+b) > 13.$

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You know that $~a \in \{2,3,5,7,11\}.$

$~a = 11~$ can be rejected, since $~a~$ must divide $~b + 13,~$ which would require $~b = 9.~$ Immediate that the ordered pair $~(a,b) = (11,9)~$ is simply too large.

You also have that $~b^2 + ba + (a^2 - 169) = 0,~$
which can be interpreted as a quadratic equation in $~b.~$

Therefore,

$$b = \frac{1}{2} \left[ -a \pm \sqrt{a^2 - 4(a^2 - 169)}\right].$$

This implies that

$$676 - 3a^2$$ is a perfect square, where $~a \in \{2,3,5,7\}.$

At this point, I am forced to resort to trial and error. Only $~a = 7,~$ results in $676 - 3a^2$ being a perfect square.

So:

• $~a = 7.~$

• $a + b > 13 \implies 6 < b < 13.$

• $a~$ divides $~b + 13 \implies b \in \{1,8\}.$

So, the only try is $~(a,b) = (7,8).$

Answer 2

Came up with something. Not sure whether its better than trial and error once you realize $a$ is a prime less than 15.

$\cos 120^\circ = -1/2$

$a^2+b^2-2ab\cos 120^\circ = 13^2$

$b^2+ab +(a^2-13^2)=0\implies b|(a^2-13^2)$

$b=\frac{-a \pm \sqrt{a^2-4(a^2-13^2)}}{2}=\frac{-a \pm \sqrt{26^2-3a^2}}{2}$

$0\le 26^2-3a^2\implies a<15$

$26^2-3a^2=p^2 \in \mathbb Z$

$p^2 \equiv 1 \pmod{3}\implies p=3k\pm 1$, $k\ge 0$ since $p>0$

$(26-p)(26+p)=3a^2$

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Case $p=3k+1$

$(25-3k)(27+3k)=3a^2$

$\implies (25-3k)(9+k)=a^2$

Either $a^2$ divides one factor and the other factor is 1 or both factors are equal to $a$.

$24-3k=a^2 \implies 9+k=1$, contradiction since $k\ge0$.

$9+k=a^2\implies k=8\implies a^2=17$, contradiction since $\sqrt{17}$ is not an integer.

It follows that $25-3k=9+k=a\implies k=4\implies a=13\implies b=-13$, contradiction.

So we can rule out $p=3k+1$

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Case $p=3k-1$

$(27-3k)(25+3k)=3a^2$

$(9-k)(25+3k)=a^2$

Again, either one of the factors is $a^2$ and the other is $1$, or both are equal to $a$.

$k\ge0 \implies 25+3k>1$ so,

$9-k=1\implies k=8 \implies a=7\implies b=8$ and we have a solution.

Suppose $9-k=25+3k\implies 4k=-16 $, contradiction since $k\ge 0$.

So there is only the one solution $a=7, b=8$.