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Say, $a=b+c$, $a$ may be rational or irrational. However, the constraint on $a$ is that $a^2$ is an integer. b>0 , c>0 which means a>0. Wanted to confirm that either $b^2$ or $c^2$ or both can't be irrational.

I reasoned like this : $a^2 =(b+c)^2$ or $a^2=b^2+2bc+c^2\tag{1}$ Since the sum of a rational and irrational number is irrational (is it correct?), and the left hand side of $(1)$ is $a^2$ which is an integer, i.e., a rational number, neither $b^2$ nor $c^2$ can be irrational. For that matter, $2bc$ can't be irrational either. Conclusion: $a^2, b^2$ and $bc$ are rational numbers.

If any of my friends in this forum can help me confirm that above conclusion is correct, I will be grateful.

Shankar1719
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    $1=\sin^2 (1)+\cos^2 (1)$. – lulu Mar 27 '22 at 11:36
  • I understand that you have a=1 , what is b ? What is c? Since, $a^2=b^2+c^2+2bc$ , in the above counter-example, what is 2bc?

    I apologize for missing these constraints : b>0 and c>0. In that case, is the conclusion valid that b^2,c^2 and bc must be rational. Can you provide some counterexamples involving non-transcedental functions? I have meanwhile edited my earlier question. Will wait for your further feedback. Thanks

     – Shankar1719 Mar 27 '22 at 13:35
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    Obviously you can do it with algebraic numbers. Just try. $1=(\sqrt[3] 2-1)+(2-\sqrt[3] 2)$. You should have no difficulty producing examples. – lulu Mar 27 '22 at 13:39
  • Great. Thanks a lot. – Shankar1719 Mar 27 '22 at 14:50

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Off the top of my head, I could think of b = 2^(1/4) and c = -2^(1/4) as counterexamples.

Alp
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