Could you help me solve this theorem? I don't understand how I can taste him.
Prove: For every a and b integers and every m positive integer, if a ≡ b (mod m), then a² ≡ b² (mod m).
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3Welcome to Mathematics Stack Exchange. Note $a^2-b^2=(a-b)(a+b)$, so, if $m\mid a-b,$ then $m\mid a^2-b^2$ – J. W. Tanner Jun 30 '21 at 02:19
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$(b+mk)^2=b^2+2bmk+m^2k^2=b^2+m(2bk+mk^2)$. – Kenta S Jun 30 '21 at 02:19
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This is a particular case of the Polynomial Congruence Rule – J. W. Tanner Jun 30 '21 at 02:30
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Cf. this question – J. W. Tanner Jun 30 '21 at 02:42
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1If $a = b + mk$ then $a^2 = (b+ mk)^2 = b^2 + 2bmk + m^2k^2$. – fleablood Jun 30 '21 at 03:01