Let $a, b$ be integers. Show that the equation $a^2 + b^2 = 26a$ has at least $12$ solutions.
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Tags: number theory, Diophantine equation, diophantine
OlympusHero
18.09.2021 16:44
We have $b^2=26a-a^2=a(26-a)$. We can see that $a=0, b=0$ works, $a=1, b=\pm 5$ works, $a=8, b=\pm 12$ works, $a=13, b=\pm 13$ works, $a=18, b=\pm 12$ works, $a=25, b=\pm 5$ works, and $a=26, b=0$ works. This is a total of $12$ solutions, as desired.
The.survivor
18.09.2021 20:14
${{a}^{2}}+{{b}^{2}}=26a\Leftrightarrow {{a}^{2}}-26a+{{b}^{2}}=0,a\in \mathbb{R}\Rightarrow {{\Delta }_{a}}\ge 0\Leftrightarrow b\le 13$
kante314
19.09.2021 03:44
We have $$a^2 + b^2 = 26a \implies a^2-26a+b^2=0$$By the quadratic formula,
$$a=\frac{-26 \pm \sqrt{676-4b^2}}{2}=-13 \pm 2 \cdot \sqrt{169-b^2}$$Thus, we want $\sqrt{169-b^2}$ to be an integer. This happens when $b=0,5,12,$ and $13$. Thus, we see the pairs are $$(a,b)=(26,0),(25,5),(1,5),(18,12),(8,12),(13,13)$$Notice having a negative $b$ wouldn't change anything, thus there are $6 \cdot 2=12$ solutions.
oralayhan
19.09.2021 17:43
Solution $(0,0)$ must be too.
numbertheory97
26.08.2024 23:44
Rewrite the equation as $(a - 13)^2 + b^2 = 169$. This gives exactly $12$ solutions for the pair $(a - 13, b)$: $(\pm 13, 0)$, $(0, \pm 13)$, $(\pm 5, \pm 12)$, and $(\pm 12, \pm 5)$. All of them give clearly distinct solutions for $(a, b)$, so we're done.