三角数tn=n(n+1)/2が平方数m^2ならば・・・

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n(n+1)=2m^2,n^2+n=2m^2,n=2m^2-n^2

t3n+4m+1=(3n+4m+1)(3n+4m+2)/2=M（平方数）

(3n+4m+1)^2+(3n+4m+1)=9n^2+16m^2+1+24nm+8m+6n+3n+4m+1

=9n^2+16m^2+2+24nm+12m+9n

=9n^2+16m^2+2+24m(2m^2-n^2)+12m+9(2m^2-n^2)　

=34m^2+2+48m^3-24nm^2+12m=2M

M=17m^2+1+24m^3-12nm^2+6m

M=17m^2+1+24m^3-12m^2(2m^2-n^2)+6m・・・nを消去できないので、この方法では証明は無理

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P＝(α^k+β^k)/2

q=(α^k-β^k)/2√2

α=3+2√2，β=3-2√2 ,α＋β=6,αβ=1

n={(α^k+β^k)/4-1/2}

m={(α^k-β^k)/4√2}

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n(n+1)={(α^k+β^k)/4-1/2}{(α^k+β^k)/4+1/2}

={(α^k+β^k)/4}^2-1/4

={(α^2k+β^2k+2)/16}-1/4

={(α^2k+β^2k-2)/16}=2m^2

2m^2=2{(α^k-β^k)/4√2}^2={(α^k-β^k)/4}^2

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3n={(3α^k+3β^k)/4-3/2}

4m={(2√2α^k-２√2β^k)/4}

3n+4m=(α^k+1+β^k+1)/4-3/2

t3n+4m+1=(3n+4m+1)(3n+4m+2)/2={(α^k+1+β^k+1)/4-1/2}{(α^k+1+β^k+1)/4+1/2}/2

={(α^k+1+β^k+1)/4}^2-1/4}/2

={(α^2k+2+β^2k+2+2)/32-1/8

={(α^2k+2+β^2k-2)/32

={(α^k+1-β^k-1)^2/32・・・惜しい

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