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Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-

\begin{align} a^2 + b^2 & = c^3 \\ (a + b)^2 & = c^4 \end{align}

Thank you for your help.

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    I feel many will be irritated by your unnecessary $\LaTeX$. – 2017-02-15
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    why , is it forbidden here ? – 2017-02-15
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    I respect that , but that what i can do – 2017-02-15
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    Your way of writing MathJax code is beyond bizarre. Did you have some software package write it for you? They usually produce code that looks as if a psychotic wrote it. I haven't seen one quite like this before. I cleaned it up. – 2017-02-15
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    The use of MathJax code (incorrectly called LaTeX by many around here) is encouraged where appropriate, but your way of doing it would certainly give MathJax a bad name if it were thought to be standard usage. – 2017-02-15
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    look, i do my best to obey system rules . i use a software to translate to this code m,i didn't learn anything about codes . i need to be a member of this community , so , what is your advice? – 2017-02-15
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    what are theses 7 solutions – 2017-02-15
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    @Moo Could you post these solutions? Of course, $(a,b,c)=(2,2,2)$ is one. Be aware that negative values are forbidden. – 2017-02-15
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    @DietrichBurde I agree, I too can only find one solution. – 2017-02-15
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    @Moo I see. I think that $a,b,c\in \mathbb{N}$ says that $a,b,c\ge 1$. So we only have $(2,2,2)$. – 2017-02-15

3 Answers 3

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Hint: Since $(a^2+b^2)c=c^4=(a+b)^2$ we have that $a^2+b^2$ divides $(a+b)^2$. This implies $a=b$ for positive integers. Then we have $2a^2=c^3$, so that $(a,b,c)=(2,2,2)$.

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First observe that $(a+b)^2=c^4 \implies a+b = c^2 , c>0$ Then we can combine that with $a^2+b^2=c^3$ to get $(a-b)^2=2c^3-c^4$. As $a-b \in N, (2c^3-c^4)^{1/2}= c(2c-c^2)^{1/2} \in N$ Which implies $c = 1,2$ . Checking the two values, $c=0$ has no natural solutions and $c=1$ has one natural solution of $a=2,b=2,c=2$.

If we include zero in the set of natural numbers then for the cases $c=0,1$ we have $(0,1,1)$ and $(1,0,1)$ as solutions. For the case $c=0$ we also have a fourth solution of $(0,0,0)$

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We have that $$ c=\frac{(a+b)^2}{a^2+b^2}=\frac{a^2+2ab+b^2}{a^2+b^2}=1+\frac{2ab}{a^2+b^2}, $$

so $\dfrac{2ab}{a^2+b^2}$ must be an integer. But by Cauchy inequality $2ab \leq a^2+b^2$ and an integer can be only if $2ab=a^2+b^2$. It implies that that $a=b$ and $c=2$. From the first equation we have $2 a^2=c^3=8 \implies a=2.$

Se we get the one solution $(2,2,2).$

If $0 \in \mathbb{N}$ then we have trivial solution $(0, 0,0).$ Also there are two more cases when $2ab \leq a^2+b^2$ is integer for $ a=0 \implies b=1$ and for $b=0 \implies a=1.$ So we get also two solutions $(0,1,1)$ and $(1,0,1).$

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    thank you , great solution , but the triples (a=0,b=0,c=0),(a=0,b=1,c=1),(a=1,b=0,c=1) also satisfing – 2017-02-15
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    well.. it depends if $0$ is a natural number. – 2017-02-15